jmc

To make such a number, we'd start with the last digit. We'd then consider different combinations of the first two digits that give us this sum. We always have $10$ choices for the third digit. $$\begin{array}{cc}\text{Last digit}& \text{First two digits}\\ 0& -\\ 1& 10\\ 2& 11,20\\ 3& 12,21,30\\ 4& 13,22,31,40\\ 5& 14,23,32,41,50\\ 6& 15,24,33,42,51,60\\ 7& 16,25,34,43,52,61,70\\ 8& 17,26,35,44,53,62,71,80\\ 9& 18,27,36,45,54,63,72,81,90\end{array}$$ The third digit can be any of the $10$ digits. The answer is $(1+2+3+4+5+6+7+8+9)\times 10=\boxed{450}.$