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} $.