jmc

Note that the only positive 2-digit palindromes are multiples of 11, namely $11,22,\dots ,99$. Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$. Then, $N=110$, and the sum of the digits of $N$ is $1+1+0=\boxed{2}$. * There are other sets of 2-digit numbers that satisfy this rule. Some of them are $110=11+33+66$ and $110=22+33+55$