Find all two-digits $N$ for which the sum of the digits of $10^N-N$ is divisible by $170$.

The digit sum of $N-1$ is somewhere between $1$ and $17$ (inclusive). So our two-digit number must necessarily be such that $9N$ is within "reach" of a multiple of $170$. The possible values for $9N$ and thus $N$ is therefore $$ \begin{array}{|c|c|c|} \hline 9N & N & \text{Digit sum}\\ \hline 171&19 & 9\\ \hline 180&20 & 10\\ \hline 342&38 & 10\\ \hline 351&39 & 11\\ \hline 513&57 & 11\\ \hline 522&58 & 12\\ \hline 684&76 & 12\\ \hline 693&77 & 13\\ \hline 855&95 & 13\\ \hline 864&96 & 14\\ \hline \end{array} $$ We see that the $N$ for which $9N$ minus the digit sum of $N-1$ is a multiple of $170$ are the larger one in each "pair", which is to say $$ 20, 39, 58, 77, 96 $$