Okay, let's see. I need to evaluate the sum S = 1/(1 + ω) + 1/(1 + ω²) + ... + 1/(1 + ω¹⁹⁹⁷), where ω is a primitive 1997th root of unity, right? Wait, actually, ω is given such that ω¹⁹⁹⁷ = 1 and ω ≠ 1, so ω is a 1997th root of unity, but not equal to 1. So, it's a primitive one? Hmm, maybe not necessarily primitive, but just a non-1 root. Anyway, the sum goes from k=1 to k=1997 of 1/(1 + ωᵏ). 

First, I remember that roots of unity have some nice properties. For example, if ω is a primitive nth root of unity, then the minimal polynomial is the nth cyclotomic polynomial, and the sum over all roots of certain expressions can be simplified. But here, ω might not be primitive, but it's a 1997th root, so n = 1997. Wait, 1997 is a prime number, right? Let me check: 1997 divided by, say, 3 is 665.666..., not integer. Divided by 7: 1997 ÷ 7 is around 285.28, not integer. 11: 1997 ÷ 11 is approximately 181.545... Also not an integer. So 1997 is prime. Therefore, the only primitive 1997th roots are those where the exponent is coprime to 1997. But ω is given as a 1997th root of unity, ω ≠ 1, so it's a primitive one if the exponent is coprime. But maybe not necessary here. 

So the sum is over k=1 to 1997 of 1/(1 + ωᵏ). Hmm. Let's think about how to evaluate such a sum. Maybe we can pair terms or find a symmetry? Since ω is a root of unity, ωᵏ is also a root of unity. For example, if we pair terms k and n - k, where n is 1997. Wait, but 1997 is odd, so n is odd. If we pair k and n - k, but since n is odd, n - k = 1997 - k, which would be different from k for all k from 1 to 1996. Then, the 1997th term would be k = 1997, which is ω¹⁹⁹⁷ = 1. So, the last term is 1/(1 + 1) = 1/2. 

Wait, so maybe the sum can be grouped into pairs where each pair is 1/(1 + ωᵏ) + 1/(1 + ω^{n - k}) where n = 1997. Then, let's compute such a pair. Let's compute 1/(1 + ωᵏ) + 1/(1 + ω^{n - k}) = [ (1 + ω^{n - k} + 1 + ωᵏ) ] / [ (1 + ωᵏ)(1 + ω^{n - k}) ] = [ 2 + ωᵏ + ω^{n - k} ] / [ 1 + ωᵏ + ω^{n - k} + ωⁿ ]. But ωⁿ = ω^{1997} = 1, so the denominator becomes 1 + ωᵏ + ω^{n - k} + 1 = 2 + ωᵏ + ω^{n - k}. Therefore, the numerator is 2 + ωᵏ + ω^{n - k}, denominator is 2 + ωᵏ + ω^{n - k}. So each pair sums to 1. 

Wait, that's interesting. So each pair of terms 1/(1 + ωᵏ) + 1/(1 + ω^{n - k}) = 1. Therefore, if we pair all terms from k = 1 to k = 1996, since 1997 is odd, we have (1997 - 1)/2 = 998 pairs, each contributing 1, and then the last term is k = 1997, which is 1/(1 + 1) = 1/2. Therefore, the total sum would be 998 * 1 + 1/2 = 998.5. But 998.5 is equal to 1997/2, which is 998.5. Wait, 1997 is 2*998 + 1, so 1997/2 = 998.5. So, the sum S = 1997/2. But hold on, let me verify this.

Wait, n = 1997, an odd prime. So, the sum from k=1 to k=n of 1/(1 + ωᵏ) where ωⁿ = 1, ω ≠ 1. Then, pairing k and n - k: since n is odd, each pair is k and n - k, which are distinct for k from 1 to (n - 1)/2. Then, the sum of each pair is 1, so the total number of pairs is (n - 1)/2, each contributing 1, and then the last term k = n, which is 1/(1 + ωⁿ) = 1/2. So, total sum is (n - 1)/2 + 1/2 = n/2. Therefore, S = n/2. But here n = 1997, so S = 1997/2. But 1997 is odd, so 1997/2 is 998.5. Hmm, but the problem states ω ≠ 1, so the sum is from k=1 to 1997, which includes the term when k=1997, which is ω^{1997} = 1, so 1/(1 + 1) = 1/2. So, the sum is (1997 - 1)/2 * 1 + 1/2 = 1996/2 + 1/2 = 998 + 0.5 = 998.5, which is 1997/2. So, is the answer 1997/2? Wait, but 1997 is a prime, but maybe there is a different approach here.

Alternatively, perhaps there is a formula for the sum over roots of unity of 1/(1 + ωᵏ). Let me recall that if ω is a primitive nth root of unity, then the sum over k=0 to n-1 of 1/(1 + ωᵏ) can be evaluated. Wait, but here we are summing from k=1 to n, which is similar but missing the term k=0. Let's check.

Wait, if we consider the sum from k=0 to n-1 of 1/(1 + ωᵏ), where ω is a primitive nth root of unity. Then, that sum is equal to n/2 when n is odd. Wait, is that a known result? Let me think. For example, when n is 3, the sum would be 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + 1). Let's compute that. For n=3, ω = e^(2πi/3). Then, 1/(1 + ω) + 1/(1 + ω²) + 1/2. Let's compute 1/(1 + ω) + 1/(1 + ω²). Since ω² is the conjugate of ω, 1/(1 + ω) + 1/(1 + ω²) = [ (1 + ω²) + (1 + ω) ] / [ (1 + ω)(1 + ω²) ] = [2 + ω + ω²] / [1 + ω + ω² + ω³]. But ω³ = 1, so denominator is 1 + ω + ω² + 1 = 2 + ω + ω². Numerator is 2 + ω + ω². So, the sum is (2 + ω + ω²)/(2 + ω + ω²) = 1. Therefore, the total sum is 1 + 1/2 = 3/2 = n/2 where n=3. So, yes, for n=3, the sum is 3/2. Similarly, for n=5, let's test. But maybe that's a general formula. If we can show that the sum from k=0 to n-1 of 1/(1 + ωᵏ) = n/2 when n is odd, then in our problem, since we are summing from k=1 to n, which is the same as from k=0 to n-1 but missing the term k=0. So, if the total sum from k=0 to n-1 is n/2, then our sum S = n/2 - 1/(1 + ω⁰) = n/2 - 1/2 = (n - 1)/2. Wait, but in the case of n=3, the sum from k=1 to 3 would be 1/(1 + ω) + 1/(1 + ω²) + 1/2. Which we saw is 1 + 1/2 = 3/2. Wait, but according to the formula above, if n=3, sum from k=0 to 2 is 3/2, which includes the term 1/(1 + 1) = 1/2. So, sum from k=1 to 3 would be 3/2 - 1/2 = 1. Wait, but that contradicts the earlier calculation where n=3 sum from k=1 to 3 was 1/(1 + ω) + 1/(1 + ω²) + 1/2 = 1 + 1/2 = 3/2. Wait, maybe my initial assumption is wrong.

Wait, perhaps the formula is different. Let me re-examine. For n=3, the sum from k=0 to 2 of 1/(1 + ωᵏ) is 3/2, as we saw. So, if we want the sum from k=1 to 3, but n=3, so k=1 to 3 is the same as k=1 to 2 and k=3. But ω^3 = 1, so the term k=3 is 1/(1 + 1) = 1/2. So, the sum from k=1 to 3 is [1/(1 + ω) + 1/(1 + ω²)] + 1/2 = 1 + 1/2 = 3/2. Wait, but according to the formula, sum from k=0 to 2 is 3/2. So, the sum from k=1 to 3 is sum from k=0 to 2 minus the term k=0, which is 1/(1 + 1) = 1/2. Therefore, 3/2 - 1/2 = 1. But that contradicts the direct computation. So, there must be a mistake here.

Wait, maybe the formula is different. Let's check the sum from k=0 to n-1 of 1/(1 + ωᵏ) when ω is a primitive nth root. For n=3, we saw it's 3/2. For n=5, let's compute. Let ω = e^(2πi/5). Then, sum_{k=0}^4 1/(1 + ωᵏ). Let's compute 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + ω³) + 1/(1 + ω⁴). The first term is 1/2. Then, similar to n=3, the sum of the other terms. Let's pair 1/(1 + ω) + 1/(1 + ω⁴) and 1/(1 + ω²) + 1/(1 + ω³). Each pair would be 1/(1 + ωᵏ) + 1/(1 + ω^{n - k}) = [2 + ωᵏ + ω^{n - k}]/[1 + ωᵏ + ω^{n - k} + 1] = [2 + ωᵏ + ω^{-k}]/[2 + ωᵏ + ω^{-k}] = 1. So, each pair sums to 1, and there are two pairs, so total sum is 1 + 1 = 2. Adding the first term 1/2, total sum is 2 + 1/2 = 5/2. Which is n/2 = 5/2. So, for n=5, the sum from k=0 to 4 is 5/2. Similarly, for n=3, sum from k=0 to 2 is 3/2. So, the general formula seems to be sum_{k=0}^{n-1} 1/(1 + ωᵏ) = n/2 when ω is a primitive nth root of unity. Therefore, if that's the case, then in our problem, we are summing from k=1 to 1997, which is the same as sum_{k=0}^{1997} 1/(1 + ωᵏ) minus the term k=0. The term k=0 is 1/(1 + 1) = 1/2. Therefore, the sum S = n/2 - 1/2 = (n - 1)/2. But in our problem, n = 1997, so S = (1997 - 1)/2 = 1996/2 = 998. But wait, earlier when n=3, sum from k=1 to 3 is 3/2, which is equal to n/2. But according to this formula, (n - 1)/2. Wait, that contradicts. Wait, in the problem statement, when n=3, if we sum from k=1 to 3, that's 3 terms, which is n=3. The formula says sum from k=0 to 2 is 3/2. Therefore, sum from k=1 to 3 is sum_{k=0}^2 - term k=0 = 3/2 - 1/2 = 1. But when we calculated directly, we saw that it was 3/2. So, that's a problem. There must be a mistake here.

Wait, maybe the formula is different. Let me check again. Let me compute for n=3. The sum from k=0 to 2 of 1/(1 + ωᵏ) is 3/2. Then, the sum from k=1 to 3 would be 3/2 minus 1/(1 + ω⁰) = 3/2 - 1/2 = 1. But when we compute 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + 1) = [1/(1 + ω) + 1/(1 + ω²)] + 1/2. We saw that [1/(1 + ω) + 1/(1 + ω²)] = 1, so total is 1 + 1/2 = 3/2. So, that contradicts the formula. Therefore, the formula must be different. So, perhaps the sum from k=0 to n-1 of 1/(1 + ωᵏ) is equal to n/2 when ω is a primitive nth root of unity. Then, in that case, the sum from k=1 to n would be sum_{k=0}^{n-1} 1/(1 + ωᵏ) minus 1/(1 + 1) = n/2 - 1/2. But in the case of n=3, that would be 3/2 - 1/2 = 1, but the actual sum is 3/2. Therefore, there must be something wrong here.

Alternatively, perhaps the formula is that the sum from k=0 to n-1 of 1/(1 + ωᵏ) is n/2 when n is even, but when n is odd, it's (n + 1)/2. Wait, let's check n=3: (3 + 1)/2 = 2, which is not 3/2. Hmm. Maybe that's not the case.

Alternatively, maybe the formula is different. Let's compute the sum more carefully. For general n, let me consider the sum S = sum_{k=0}^{n-1} 1/(1 + ωᵏ). Let's consider multiplying numerator and denominator by (1 - ωᵏ):

1/(1 + ωᵏ) = (1 - ωᵏ)/(1 - ω^{2k}) if ω^{2k} ≠ 1. But since ω is a primitive nth root, 1 - ω^{2k} ≠ 0 unless ω^{2k} = 1, which would require 2k ≡ 0 mod n. Since n is prime (1997), this would only happen if k ≡ 0 mod n/2, but n is odd, so 2 and n are coprime. Therefore, 2k ≡ 0 mod n implies k ≡ 0 mod n. But k is from 0 to n-1, so only k=0 satisfies that. Therefore, for k ≠ 0, we can write 1/(1 + ωᵏ) = (1 - ωᵏ)/(1 - ω^{2k}). Then, sum S = sum_{k=0}^{n-1} 1/(1 + ωᵏ) = 1/(1 + 1) + sum_{k=1}^{n-1} (1 - ωᵏ)/(1 - ω^{2k}) = 1/2 + sum_{k=1}^{n-1} (1 - ωᵏ)/(1 - ω^{2k}).

But this seems complicated. Alternatively, perhaps consider that the sum S satisfies (1 + ωᵏ)S = sum_{k=0}^{n-1} 1 = n. Wait, but that's not directly helpful.

Alternatively, consider that 1/(1 + ωᵏ) + 1/(1 + ω^{n - k}) = [ (1 + ω^{n - k}) + (1 + ωᵏ) ] / [ (1 + ωᵏ)(1 + ω^{n - k}) ] = [ 2 + ωᵏ + ω^{n - k} ] / [ 1 + ωᵏ + ω^{n - k} + ωⁿ ]. But ωⁿ = 1, so denominator becomes 2 + ωᵏ + ω^{n - k}. Therefore, numerator and denominator are equal, so each pair sums to 1. Thus, if we pair k and n - k, each pair sums to 1. Since n is odd, we have (n - 1)/2 pairs, each contributing 1, and the term k = (n + 1)/2 is unpaired? Wait, n=1997, which is odd. Let's see, for n=1997, k ranges from 0 to 1996. Wait, in the problem, the sum is from k=1 to 1997, but in the formula, it's from k=0 to 1996. So, perhaps there is confusion here. Let me clarify.

Wait, in the problem statement, we are given ω^{1997} = 1 and ω ≠ 1. Then, the sum is 1/(1 + ω) + 1/(1 + ω²) + ... + 1/(1 + ω^{1997}). Wait, but ω^{1997} = 1, so the last term is 1/(1 + 1) = 1/2. But the exponents in the sum go from 1 to 1997. However, since ω^{1997} = 1, the exponents are modulo 1997. So, the exponents are from 1 to 1997, which is equivalent to exponents from ω^1 to ω^{1997} = ω^0 = 1. So, the sum is from k=1 to 1997 of 1/(1 + ω^k) = sum_{k=1}^{1997} 1/(1 + ω^k). But since ω^{1997} = 1, the term k=1997 is 1/(1 + 1) = 1/2. The other terms are 1/(1 + ω^k) for k=1 to 1996.

So, in effect, the sum is sum_{k=1}^{1997} 1/(1 + ω^k) = sum_{k=1}^{1996} 1/(1 + ω^k) + 1/(1 + 1). 

Now, since ω is a primitive 1997th root of unity, then ω^k for k=1 to 1996 are all the non-1 roots. So, the sum S = sum_{k=1}^{1996} 1/(1 + ω^k) + 1/2.

Now, if we can compute sum_{k=1}^{1996} 1/(1 + ω^k), then add 1/2 to get the total. 

But, as before, when we pair k and 1997 - k, each pair sums to 1. Since 1997 is odd, 1997 - k is different from k for k=1 to 1996. Therefore, the number of pairs is (1996)/2 = 998. Each pair contributes 1, so the sum sum_{k=1}^{1996} 1/(1 + ω^k) = 998 * 1 = 998. Therefore, the total sum S = 998 + 1/2 = 998.5 = 1997/2. 

Therefore, the answer should be 1997/2. 

But let me verify this with a smaller prime, say n=3. If n=3, then sum_{k=1}^{3} 1/(1 + ω^k). Let's compute this. The roots are ω^1, ω^2, ω^3=1. So, the sum is 1/(1 + ω) + 1/(1 + ω²) + 1/2. As we saw earlier, 1/(1 + ω) + 1/(1 + ω²) = 1. So, total sum is 1 + 1/2 = 3/2, which is 3/2 = n/2. Wait, but n=3, so 3/2 is indeed n/2. But according to the previous logic, sum_{k=1}^{n} 1/(1 + ω^k) = (n - 1)/2 + 1/2 = n/2. So, in the case of n=3, sum is 3/2 = 3/2. Which matches. Similarly, for n=5, sum is 5/2. So, yes, it seems that for prime n, the sum from k=1 to n of 1/(1 + ω^k) = n/2. Therefore, in the problem, n=1997, so the answer is 1997/2. 

Therefore, the final answer is \boxed{\dfrac{1997}{2}}.
</think>