Let $N = \overline{a_na_{n-1} \ldots a_0}$ and $m$ be a positive integer with $m \geq n$. We may assume that $a_0 \neq 0$ since adding a zero doesn't change the digit sum. Then \begin{align*}N(10^m-1) &= (10^n a_n+\cdots+a_0)(10^m-1)\\&= (10^{m+n}a_n+\cdots+10^ma_0)-(10^n a_n+\cdots+a_0)\\ &=10^{m+n}a_n+\cdots+(a_0-1)10^m+9(10^{m-1}+\cdots+10^n)+(9-a_{n-1})10^{n-1}+(9-a_{n-2})10^{n-2}+\cdots+(10-a_0).\end{align*}Thus, $s(N(10^m-1)) = s(10^m-1) = 9m$ where $s(k)$ denotes the sum of the digits of a positive integer $k$.