6. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits be denoted as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". The sum of all coincidence numbers is $\qquad$ .
6.531.

Let $n=\overline{a_{1} a_{2} \cdots a_{k}}\left(a_{1} \neq 0\right)$.
From $n-s(n)=p(n)$, we get
$$
\sum_{i=1}^{k-1} a_{i}\left(10^{k-i}-1\right)=a_{1} a_{2} \cdots a_{k} \text {, }
$$

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.
Thus, $k=2$.
Hence, we can set $n=\overline{a_{1} a_{2}}=10 a_{1}+a_{2}$. From
$$
a_{1}+a_{2}+a_{1} a_{2}=10 a_{1}+a_{2} \text {, }
$$

we get $9 a_{1}=a_{1} a_{2}$.
Therefore, $a_{2}=9, a_{1} \in\{1,2, \cdots, 9\}$, which means all coincidental numbers are $19,29, \cdots, 99$, and their sum is 531.