Suppose that $( n,m)$ satisfies the condition. Note that the least proper divisor is always prime number. Let $q$ be the least proper divisor of $n$. Then $q+1$ is the least proper divisor of $m$. In particular, both $q$ and $q+1$ are prime numbers, which implies $q=2$. Since $n/2$ is the largest proper divisor of $n$, $n/2+1$ is the largest proper divisor of $m$. So we have $n/2+1=m/3$. Since $2\nmid m$, we must have $4\mid n$. So we can take $k\in \mathbb{N}$ such that $n=4k$. Thus we have $m=3( 2k+1)$. If $k=1$, then $( n,m) =( 4,9)$, which satisfies the condition indeed. Suppose that $k\geq 2$. Since $k$ is a proper divisor of $n$, we have $k+1\mid 3( 2k+1)$. It follows that $k=2$, which implies $n=8$. We can verify that $( n,m) =( 8,15)$ satisfies the condition indeed. Hence the answer is $n=4,8$.