A $\textit{palindrome}$ is an integer that reads the same forwards and backwards. How many positive 3-digit palindromes are multiples of $3$?
A $3$-digit palindrome must be of the form $aba$, where $a$ and $b$ are digits, and $a\neq 0$. In order for $aba$ to be divisible by $3$, we require that $a + b + a = 2a + b$ be divisible by $3$. Since $0 < a\leq 9$ and $0 \leq b \leq 9$, the maximum possible value of $2a+b$ is $2\cdot 9 + 9 = 27$.  We will list all of the multiples of $3$ from $0$ through $27$, and determine how many possibilities for $a, b$ make $2a + b$ equal to that multiple.

If $2a + b = 0$, then there are no solutions such that $a \neq 0$.
If $2a+b=3$, then $b=3-2a$, so $a=1$ is the only solution.
If $2a+b=6$, then $b=6-2a$, so $a=1,2,3$, since $a\ge 4$ will make $b$ negative.
If $2a+b=9$, then $b=9-2a$, so $a=1,2,3,4$, since $a\ge 5$ will make $b$ negative.
If $2a+b=12$, then $b=12-2a$, so $a=2,3,4,5,6$, since $a\le 1$ will make $b\ge 10$, and $a\ge 7$ will make $b$ negative.
If $2a+b=15$, then $b=15-2a$, so $a=3,4,5,6,7$, since $a\le 2$ will make $b\ge 10$, and $a\ge 8$ will make $b$ negative.
If $2a+b=18$, then $b=18-2a$, so $a=5,6,7,8,9$, since $a\le 4$ will make $b\ge 10$, and $a$ must be less than $10$.
If $2a+b=21$, then $b=21-2a$, so $a=6,7,8,9$, since $a\le 5$ will make $b\ge 10$, and $a$ must be less than $10$.
If $2a+b=24$, then $b=24-2a$, so $a=8,9$, since $a\le 7$ will make $b\ge 10$, and $a$ must be less than $10$.
If $2a+b=27$, then $b=27-2a$, so $a=9$, since as we've seen $a$ and $b$ must both be as large as possible.

In each case, a value of $a$ uniquely determines a value of $b$, so we haven't missed any palindromes. Thus the total number is $1+3+4+5+5+5+4+2+1=\boxed{30}$.