Five. (Full marks 14 points) If the sum of the digits of a natural number and the product of its digits add up to exactly the natural number itself, we call it a "lucky number." Try to find the sum of all "lucky numbers."

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Let the $n$-digit natural number be $N=\overline{a_{1} a_{2} \cdots a_{n}}$, then 
\[ N=a_{1} \times 10^{n-1}+a_{2} \times 10^{n-2}+\cdots+a_{n-1} \times 10+a_{n} \]
(Where $a_{1}, a_{2}, \cdots, a_{n}$ are the digits of $N$).
If $N$ is a "lucky number", then
\[ 
\begin{array}{l}
a_{1} \times 10^{n-1}+a_{2} \times 10^{n-2}+\cdots+a_{n-1} \times 10+a_{n} \\
=\left(a_{1}+a_{2}+\cdots+a_{n}\right)+a_{1} \times a_{2} \times \cdots \times a_{n},
\end{array}
\]

Rearranging gives
\[ 
\begin{array}{l}
a_{1}\left(10^{n-1}-1\right)+a_{2}\left(10^{n-2}-1\right)+\cdots+a_{n-1}(10-1) \\
=a_{1} \times a_{2} \times \cdots \times a_{n},
\end{array}
\]

That is, 
\[ a_{1}(\underbrace{99 \cdots 9}_{n-14}-a_{2} \times a_{3} \times \cdots \times a_{n})+A=0, \]
where $A=a_{2}\left(10^{n-1}-1\right)+\cdots+a_{n-1}(10-1)$ is a non-negative number.
When $n \geqslant 3$, $\underbrace{99 \cdots 9}_{n-14}>9^{n-1} \geqslant a_{2} \times a_{3} \times \cdots \times a_{n}$, hence (1) does not hold, so there are no "lucky numbers" with 3 or more digits;
When $n=2$, from (1) we get $a_{1}\left(9-a_{2}\right)=0$,
Since $a_{1} \neq 0, \therefore 9-a_{2}=0$, thus $a_{2}=9$.
Therefore, two-digit numbers ending in 9 are "lucky numbers" (it can be verified that no single-digit numbers are "lucky numbers"), they are $19,29,39,49$, $59,69,79,89,99$ and their sum is 531.