12. (2004 Czech and Slovak Mathematical Olympiad) Find the positive integer $n$, such that $\frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{n!}$ is an integer.
12. When $n=1,2,3$, the sums are $1,3,5$, respectively, which are integers.

When $n>3$,
$$
\begin{aligned}
& \frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{(n-2)!}+\frac{n}{(n-1)!}+\frac{n}{n!} \\
= & \frac{n(n-1) \cdots 2+n(n-1) \cdots 3+\cdots+n(n-1)+n+1}{(n-1)!} .
\end{aligned}
$$

For it to be an integer, the numerator must be divisible by $n-1$, so $n+1$ must be divisible by $n-1$, which means $2[=n+1-(n-1)]$ must be divisible by $n-1$. Therefore, $n-1 \in\{1,2\}$, which is impossible.
Thus, the sum is an integer if and only if $n \in\{1,2,3\}$.