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
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 simplifies to \( a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right) + m = 0 \),
where \( m = a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \geqslant 0 \).
If \( k \geqslant 3 \), since
\[
10^{k-1}-1-a_{2} a_{3} \cdots a_{k} \geqslant 10^{k-1}-1-9^{k-1} > 0,
\]

this 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
\[
9 a_{1} = a_{1} a_{2} \Rightarrow a_{2} = 9 \left(a_{1} \in \{1, 2, \cdots, 9\}\right),
\]

so the complete set of coincidental numbers is \( 19, 29, \cdots, 99 \), and their sum is 531.