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 $n \in\{1,2,3\}$.