Clearly $n>0$, and $n=3k$ since otherwise those two divisors cannot be both integers. We have that $k+a, k-a$ both divide $3k$, for some positive integer $a$, where $1\leq a<k$. Note that $k+a$ is a divisor of $n$ greater than $k$, so it may only have two values: $\frac{n}{2}$ and $n$. Since $a<k$, we have $k+a<2k<n$, so it must be the case that $k+a=\frac{n}{2}$. This implies $n$ is even. Now write $n=6m$; we need $k+a=2m+a=3m$, i.e $a=m$. This way $k+a=3m$ is always a divisor of $n$, and $k-a=2m-m=m$ is also always a divisor. Hence the numbers we are looking for are all positive multiples of 6.