surprising, it's not hard! Assume that $p_k$ is a maximal of $\{p_1,\ldots,p_k\}$. So we have $p_k\mid p_j+1$ for some $j$. If $p_k>3$ then $p_j>2$ and then $p_j+1$ must be a composite! So $p_k<p_j+1$ so $p_k\leq p_j$, which implies $p_k=p_j$, contract to the fact $p_k\mid p_j+1$. So $p_k=3$. Assume we have $r\geq0$ numbers $p_j$ which equal to $2$ and $s=k-r\geq0$ ones equal to $3$. So we have $2^r3^{s}\mid 3^r\cdot 4^s$. So we have $r\leq 2s$ and $s\leq r$. Answer $n=2^r3^s$ where $0\leq s$ $\leq r\leq 2s$.