8. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits as $p(n)$. If $s(n)+$ $p(n)=n$ holds, then $n$ is called a "coincidence number". Therefore, the sum of all coincidence numbers is $\qquad$.
8.531.

Let $n=\overline{a_{1} a_{2} \cdots a_{k}}\left(a_{1} \neq 0\right)$.
From $n-s(n)=p(n)$, we get
$$
\begin{array}{l}
a_{1}\left(10^{k-1}-1\right)+a_{2}\left(10^{k-2}-1\right)+\cdots+a_{k-1}(10-1) \\
=a_{1} a_{2} \cdots a_{k},
\end{array}
$$

which means $a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right)+m=0$,
where $m=a_{2}\left(10^{k-2}-1\right)+\cdots+a_{k-1}(10-1) \geqslant 0$.
If $k \geqslant 3$, since
$$
10^{k-1}-1-a_{2} a_{3} \cdots a_{k} \geqslant 10^{k-1}-1-9^{k-1}>0 \text {, }
$$

it contradicts equation (1). Therefore, $k \leqslant 2$.
When $k=1$, $n=s(n)$, which does not meet the condition, so $k=2$.
Thus, we can set $n=\overline{a_{1} a_{2}}=10 a_{1}+a_{2}$.
Combining $a_{1}+a_{2}+a_{1} a_{2}=10 a_{1}+a_{2}$, we get
$$
9 a_{1}=a_{1} a_{2} \Rightarrow a_{2}=9\left(a_{1} \in\{1,2, \cdots, 9\}\right),
$$

which means all coincidental numbers are $19,29, \cdots, 99$, and their sum is 531.