9. $14 N$ is a 5-digit number composed of 5 different non-zero digits, and $N$ is equal to the sum of all three-digit numbers formed by taking 3 different digits from these 5 digits. Find all such 5-digit numbers $N$.
(6th Latin American Mathematical Olympiad, 1991)
[Solution] Let the 5-digit number $N=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}$.
When $a_{1}$ is in the hundreds place of a three-digit number, there are $P_{4}^{2}=12$ such numbers,
When $a_{1}$ is in the tens place of a three-digit number, there are $P_{4}^{2}=12$ such numbers,
When $a_{1}$ is in the units place of a three-digit number, there are $P_{4}^{2}=12$ such numbers.
Thus, according to the problem, we have
$$
\begin{aligned}
N & =\overline{a_{1} a_{2} a_{3} a_{4} a_{5}} \\
& =\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)(100 \cdot 12+10 \cdot 12+12) \\
& =1332\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right) .
\end{aligned}
$$

Since 9 divides 1332, then
$$
9 \mid N=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}} .
$$

Thus,
$$
91\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right) .
$$

Therefore, $N$ must be a multiple of $1332 \cdot 9=11988$.
Since
$$
\begin{aligned}
15=1+2+3+4+5 \leqslant & a_{1}+a_{2}+a_{3}+a_{4}+a_{5} \\
& \leqslant 9+8+7+6+5=35,
\end{aligned}
$$

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$ can only be 18 or 27.
(1) When $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=18$,
$$
\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}=1332 \cdot 18=23976 \text {, }
$$

However, $\quad 2+3+9+7+6=27 \neq 18$.
(2) When $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=27$,

At this time,
$$
3+5+9+6+4=27 .
$$

Therefore, the required five-digit number is only 35964.