4A. Find all natural numbers $n$ such that $\sum_{k=1}^{n} \frac{n}{k!}$ is an integer.
Solution. Since $\sum_{k=1}^{n} \frac{1}{k!}=\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}=\frac{n!+n(n-1) \ldots .3+\ldots+n(n-1)+n+1}{n!}=\frac{(n-1) S+n+1}{n!}$, we get that $\sum_{k=1}^{n} \frac{n}{k!}=\frac{(n-1) S+(n+1)}{(n-1)!}$. For the number $\sum_{k=1}^{n} \frac{n}{k!}$ to be an integer, it must be that $(n-1)!\mid(n-1) S+n+1$. Since $(n-1) \mid(n-1)!$ and $n-1 \mid(n-1) S$, it follows that $n-1 \mid n+1$. Clearly $n-1 \mid n-1+2$ so from there $n-1 \mid 2$. Therefore, $n=2$ and $n=3$.