Product $\times$ Sum $=$ Division!

Let $n = 10a+b$ , where $a=1,2,3...9$ and $b=0,1,2...9$ . Then we have:

$\begin{aligned} s(n) + p(n) & = n \\ \implies a+b + ab & = 10a+b \\ ab & = 9a \\ \implies b & = 9 \end{aligned}$

$\implies n = \{19, 29, 39, ... 99\}$ and the sum of all $n$ 's,

$S = \displaystyle \sum_{a=1}^9 (10a+9) = 10 \cdot \frac {9(10)}2 + 9\cdot 9 = 450 +81 = \boxed{531}$ .