If $p$ is smallest prime which divide $n$ then $p$ is smallest and $\frac{n}{p}=q$ is largest proper divisor so $\frac{p+q}{2} | pq$ or $p+q| 2pq \to p+q|2p(p+q)-2pq=2p^2$ $p+q \geq 2p$ so $p+q=2p, p^2$ or $p+q=2p^2$ $q=p \to n=p^2$ has $1$ proper divisor $q=p^2-p$ then $p-1|n$ but as $p$ is smallest prime, then $p=2$ and $q=2,n=4$ If $q=2p^2-p=p(2p-1)$ then $2p-1$ is prime as it can not have prime divisors less than $p$ $n=p^2(2p-1)$ has $4$ proper divisors