Okay, let me try to figure this out. The problem says that the sum of the first N positive odd numbers is 121, and we need to find N. Hmm, I remember that there's a formula for the sum of the first N odd numbers. Let me recall... I think the sum of the first N odd numbers is N squared. Wait, is that right? Let me check with a small example. If N is 1, the sum is 1, which is 1². If N is 2, the sum is 1 + 3 = 4, which is 2². For N=3, 1+3+5=9, which is 3². Yeah, that seems correct. So the formula is Sum = N². 

So if the sum is 121, then N² = 121. To find N, I just take the square root of 121. Let me calculate that. The square root of 121 is 11 because 11*11=121. Therefore, N should be 11. 

Wait, but let me make sure I didn't skip any steps here. The problem states "the first N positive odd numbers." The nth odd number is given by 2n - 1, right? So the first N odd numbers would be 1, 3, 5, ..., (2N-1). The sum of an arithmetic series is (number of terms)/2 * (first term + last term). So in this case, the number of terms is N, the first term is 1, and the last term is 2N - 1. So the sum should be N/2 * (1 + (2N - 1)) = N/2 * 2N = N². Yep, that matches the formula I remembered earlier. 

So using that formula, setting N² = 121, solving for N gives N = 11. Therefore, the answer should be 11. 

But just to double-check, let's compute the sum manually for N=11. The 11th odd number is 2*11 -1 = 21. So the sum is 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21. Let me add these up. 

1+3=4, 4+5=9, 9+7=16, 16+9=25, 25+11=36, 36+13=49, 49+15=64, 64+17=81, 81+19=100, 100+21=121. Yes, that adds up to 121. So that confirms it. 

Therefore, after verifying both through the formula and manually adding the terms, the value of N is indeed 11.
Therefore, the value of $N$ is $\boxed{11}$.
</think>