Solution:
We claim that the largest integer is $4$. We first prove that all strings $S$ of $2025$ letters contain a palindromic substring of length $4$ or longer, which implies that $k \geq 4$. Then we shall provide a construction to show that $k$ cannot be $5$ or more.

We first begin by proving $k \geq 4$. Let $R$ be the $2023$ letter substring of $S$ with the first and last letter removed.

Case 1: $R$ contains a letter that is repeated $4$ or more times in a row.
This repeating letter (e.g. $AAAA$) is a palindromic substring of length $4$ or more, so we are done.

Case 2: $R$ contains a letter that is repeated $3$ times in a row, and no more.
Without loss of generality, assume this repeated letter is $A$. As it repeats only $3$ times in a row, the letters immediately to the left and right of the substring $AAA$ must be $B$, therefore $S$ contains the substring $BAAAB$, which is palindromic with length $5 > 4$.

Case 3: $R$ contains a letter that is repeated $2$ times in a row, and no more.
Without loss of generality, assume this repeated letter is $A$. As it repeats only $2$ times in a row, the letters immediately to the left and right of the substring $AA$ must be $B$, therefore $S$ contains the substring $BAAB$, which is palindromic with length $4$.

Case 4: $R$ does not contain a letter that is repeated in a row.
As no letter repeats, $R$ must alternate between $A$ and $B$. Therefore it contains the substring $ABABA$, which is palindromic with length $5 > 4$.

This proves that $k \geq 4$.

We now provide a construction that does not contain palindromic substrings of length $5$ or more. Consider the following sequence of length $2025$ built from repeating the $6$ letters $AABABB$.

AABABB AABABB AABABB...

There are only $6$ possible five letter substrings as the sequence repeats every $6$ letters, none of these substrings are palindromic.

AABAB, ABABB, BABBA, ABBAA, BBAAB, BAABA

Similarly the six letter substrings are all not palindromic either.

AABABB, ABABBA, BABBAA, ABBAAB, BBAABA, BAABAB

Any longer palindromic substring of odd length must contain a palindromic substring with $5$ letters, and any longer palindromic substring of even length must contain a palindromic substring with $6$ letters. As we proved earlier there are no such substrings, there also cannot be palindromic substrings of length $7$ or longer in our constructed sequence.