smc

For each digit $a=1,2,\dots ,9$ there are $10\cdot 10$ (ways of choosing $b$ and $c$) palindromes. So the $a$s contribute $(1+2+\cdots +9)(100)(10^4+1)$ to the sum. For each digit $b=0,1,2,\dots ,9$ there are $9\cdot 10$ (since $a\ne 0$) palindromes. So the $b$s contribute $(0+1+2+\cdots +9)(90)(10^3+10)$ to the sum. Similarly, for each $c=0,1,2,\dots ,9$ there are $9\cdot 10$ palindromes, so the $c$ contributes $(0+1+2+\cdots +9)(90)(10^2)$ to the sum. It just so happens that $$(1+2+\cdots +9)(100)(10^4+1)+(1+2+\cdots +9)(90)(10^3+10)+(1+2+\cdots +9)(90)(10^2)=49500000$$ so the sum of the digits of the sum is $\boxed{18}$.