<Question>
Palindromes are numbers that read the same backwards and forwards, like 5665. What is the least possible positive four-digit palindrome that is divisible by 3?
</Question>

<Plan_1>
Alright, let's try to figure out the smallest four-digit palindrome that's divisible by 3. Hmm, okay, so first, I need to remember what a palindrome is.
A palindrome is a number that reads the same forwards and backwards. For a four-digit number, that means the first digit is the same as the last digit, and the second digit is the same as the third digit. So, a four-digit palindrome would look like ABBA, where A and B are digits.

Now, the question is asking for the least possible positive four-digit palindrome divisible by 3. So, the smallest four-digit number is 1000, but that's not a palindrome.
The smallest four-digit palindrome would be 1001, right? Because 1001 reversed is 1001.

But wait, let me check.
If A is 1, then the number is 1BB1. The smallest four-digit palindrome starting with 1 would have B as 0, so 1001.

Then 1111, 1221, etc.

But is 1001 divisible by 3?
To check divisibility by 3, the sum of its digits must be a multiple of 3. Let's add the digits of 1001: 1 + 0 + 0 + 1 = 2. 2 is not divisible by 3, so 1001 isn't.

So the next palindrome would be 1111.

<prompt_1.1> Check if 1111 and 1221 are divisible by 3. </prompt_1.1>

But wait, is there a smaller four-digit palindrome between 1001 and 1221 that I might have missed?

Wait, after 1001, the next palindromes would be 1111, 1221, 1331, etc.

<prompt_1.2> Verify the sequence of four-digit palindromes between 1001 and 1221. </prompt_1.2>
-----

But just to be thorough, let's check another way. Let's recall that a four-digit palindrome is of the form ABBA, which is 1000*A + 100*B + 10*B + A = 1001*A + 110*B.

<prompt_1.3> Explore the divisibility properties of four-digit palindromes. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's check the sum: 1 + 1 + 1 + 1 = 4. Still not divisible by 3. Next is 1221. Sum is 1 + 2 + 2 + 1 = 6. Oh, 6 is divisible by 3. So 1221 is a candidate.
</execution_1.1>

<execution_1.2>
, but maybe if we increment the middle digits first. Wait, how exactly do four-digit palindromes increment? Let's think. Since it's ABBA, increasing the number would involve first increasing the middle digits. For example, 1001, then 1111, then 1221, 1331, ...., 1991, then 2002, and so on. So between 1001 and 1221, the palindromes are 1001, 1111, 1221. So yes, those are the only ones.
</execution_1.2>

<execution_1.3>
So, 1001A + 110B. We can factor that as 11*(91A + 10B). So, any four-digit palindrome is divisible by 11. Wait, that's interesting. So, 1221 is 11*111, which is 11* (91*1 + 10*2) = 11*(91 + 20) = 11*111. So, since all four-digit palindromes are divisible by 11, and we need one that is also divisible by 3. Therefore, the palindrome must be divisible by the least common multiple of 11 and 3, which is 33.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
So 1221 is the third four-digit palindrome. But maybe there's another one with a higher first digit but lower overall value?
No, because the first digit is the thousands place, so higher first digits would make the number larger. So 1001 is the smallest, then 1111, then 1221.

But 1221 is divisible by 3.

So is that the answer? Wait, but let's verify.

<prompt_2.1> Verify if 1221 is divisible by 3. </prompt_2.1>
-----

Based on execution_1.3:
So, we need to find the smallest four-digit palindrome divisible by 33.

Alternatively, since the palindrome is divisible by 11 and 3, it must be divisible by 33.

<prompt_2.2> Use modular arithmetic to determine divisibility conditions for palindromes. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's check 1221 divided by 3. 3 times 407 is 1221. Yes, 3*407=1221. So that works.
</execution_2.1>

<execution_2.2>
So, perhaps looking for the smallest four-digit multiple of 33 that is a palindrome. But maybe that approach is more complicated. Alternatively, since we already know that four-digit palindromes are divisible by 11, we can just check which one is also divisible by 3.

But given that the palindrome is 11*(91A + 10B), then 91A + 10B must be divisible by 3 because 11 is prime and doesn't have 3 as a factor. Therefore, 91A + 10B ≡ 0 mod 3.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But wait, let's check if there's a smaller palindrome between 1001 and 1221 that is divisible by 3.
The only ones in that range are 1001, 1111, and 1221. We already checked 1001 and 1111 aren't divisible by 3. So 1221 must be the answer.

But wait a second, maybe there's a four-digit palindrome starting with 1 but with different middle digits. Let me confirm.

<prompt_3.1> Confirm the sequence of palindromes and their divisibility by 3. </prompt_3.1>
-----

Based on execution_2.2:
Let's compute 91 mod 3.

<prompt_3.2> Calculate the modular equivalences of 91 and 10 mod 3. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
For example, 1001, 1111, 1221, 1331, ...., 1991. Each time, the middle two digits increase by 11. So the next palindrome after 1001 would indeed be 1111, then 1221, etc. So there are no four-digit palindromes between 1001 and 1221 except 1111. Therefore, 1221 is the first one after 1001 that's divisible by 3.
</execution_3.1>

<execution_3.2>
91 divided by 3 is 30*3=90, so remainder 1. So 91 ≡ 1 mod 3. Similarly, 10 ≡ 1 mod 3. Therefore, 91A + 10B ≡ A + B mod 3. So for the entire palindrome to be divisible by 3, A + B must ≡ 0 mod 3. Therefore, the sum of the first and second digits must be divisible by 3.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Alternatively, maybe starting with a different first digit? Wait, the first digit has to be at least 1, so 1 is the smallest.

<prompt_4.1> Confirm if 1221 is the smallest four-digit palindrome divisible by 3. </prompt_4.1>
-----

Based on execution_3.2:
Ah, that's a key insight! So, instead of checking each palindrome, we can use the fact that A + B must be a multiple of 3.

<prompt_4.2> Determine the smallest palindrome using the condition A + B ≡ 0 mod 3. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
So the thousands digit is 1. So the smallest possible four-digit palindrome is 1001. Then next is 1111, then 1221. So 1221 is the third one, and the first one divisible by 3. Therefore, 1221 is the answer.
</execution_4.1>

<execution_4.2>
Since A is the first digit (from 1 to 9) and B is the second digit (from 0 to 9), we can look for the smallest four-digit palindrome (ABBA) where A + B is divisible by 3.

So, starting with the smallest A, which is 1, we need to find the smallest B such that 1 + B is divisible by 3. The possible B values would be 2, 5, 8, since 1 + 2 = 3, 1 + 5 = 6, 1 + 8 = 9, all divisible by 3. So the smallest B is 2. Therefore, the palindrome would be 1 2 2 1, which is 1221. That's the smallest B possible for A=1. So that would be the answer. Since 1221 is a four-digit palindrome, and the smallest possible when A=1 and B=2, which is the minimal B that satisfies the divisibility condition.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
But let's check if there is a smaller four-digit palindrome with A=1 and a lower B, but still A + B divisible by 3.

<prompt_5.1> Verify the smallest B value for A=1 that satisfies divisibility by 3. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
If B=2 is the smallest, then lower Bs would be 0 or 1. For B=0: 1 + 0 = 1, not divisible by 3. B=1: 1 + 1 = 2, not divisible by 3. So B=2 is indeed the smallest. Therefore, 1221 is the answer.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But let me just verify this with another approach. Let's list the four-digit palindromes in order starting from 1001 and check their divisibility by 3:

<prompt_6.1> Verify the answer using different methods and confirm 1221 as the solution. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
1. 1001: digits sum to 1+0+0+1=2, not divisible by 3.
2. 1111: digits sum to 1+1+1+1=4, not divisible by 3.
3. 1221: digits sum to 1+2+2+1=6, which is divisible by 3. So 1221 is the third four-digit palindrome and the first one that's divisible by 3. Hence, the answer should be 1221.

Therefore, after analyzing through different methods—direct checking, mathematical formulation, and modular arithmetic—it's clear that 1221 is the least four-digit palindrome divisible by 3.
</execution_6.1>
</Execution_6>

<Final_answer>
Therefore, the least possible positive four-digit palindrome that is divisible by $3$ is $\boxed{1221}$.
</Final_answer>