Note that $n=1$ works. If $n>1$, let the divisors of $n$ be $1=d_1<d_2<\dots<d_k=n$. So $d_1d_2\dots d_k \mid (d_1+1)(d_2+1)\dots(d_k+1)$. But $d_k+1=n+1$ is relatively prime with all divisors of $n$, so $d_2\dots d_k \mid (d_1+1)(d_2+1)\dots(d_{k-1}+1)$. But $d_{i +1}\ge d_i +1$ for all $i =1,2,\dots, k-1$, so $d_2\dots d_k \ge (d_1+1)(d_2+1)\dots(d_{k-1}+1)$. Thus, $d_{i +1} = d_i +1$ for all $i =1,2,\dots, k-1$, and $n=d_k=d_{k-1}+1\iff d_{k-1}=n-1$. So $n-1 \mid n \iff n-1 \mid 1 \iff n=2$. Note that $n=2$ also works. Thus, $n$ can only be either $1$ or $2$.