jmc

Once we've picked the first two digits of a $4$-digit palindrome, the last two digits are automatically chosen by mirroring the first two. Thus, we can make exactly one $4$-digit palindrome for every $2$-digit number. For example, the $2$-digit number $57$ gives the palindrome $5775$.

For an integer to be divisible by $3$, the sum of its digits must also be divisible by $3$. A $4$-digit palindrome has two identical pairs of digits. If the total of all four digits is a multiple of $3$, then the first two digits must also add up to a multiple of $3$ (since doubling a non-multiple of $3$ can't give us a multiple of $3$). Thus, to make a $4$-digit palindrome which is a multiple of $3$, we must use a $2$-digit number that is a multiple of $3$.

This tells us that the number of $4$-digit palindromes that are divisible by $3$ is identical to the number of multiples of $3$ from $10$ through $99$. Here is a list of those multiples of $3$: $$12,15,18,21,24,\dots ,90,93,96,99.$$ This list consists of the $30$ positive multiples of $3$ greater than $10.$ So, there are $30$ numbers in the list, and therefore $\boxed{30}$ four-digit palindromes that are divisible by $3$.

Here is a list of those palindromes: $$1221,1551,1881,2112,2442,\dots ,9009,9339,9669,9999.$$