Divisibility with floor

Let $m=\left\lfloor \sqrt{n} \right\rfloor$ . Since $n>4$ , then $m\ge 2$ is an integer.

We have ${{m}^{2}}\le n<{{\left( m+1 \right)}^{2}}$ , so ${{m}^{2}}\le n\le {{m}^{2}}+2m$ .

Set $n={{m}^{2}}+k$ , with $0\le k\le 2m$ .

From $m+1\mid{{m}^{2}}+k-1$ we get $m+1\mid k$ . On the other hand $k<2\left( m+1 \right)$ , thus $k=0$ or $k=m+1$ .

If $k=0$ , from $m-1\mid{{m}^{2}}+1$ follows $m-1\mid2$ , so $m=2$ or $m=3$ , hence $n=4$ or $n=9$ .

If $k=m+1$ , from $m-1\mid{{m}^{2}}+m+2=\left( m-1 \right)\left( m+2 \right)+4$ follows $m-1\mid4$ . We obtain $m=2$ or $m=3$ or $m=5$ , hence $n=7$ or $n=13$ or $n=31$ .

So the sum of all possible value of $n$ is $4+7+9+13+31=\boxed{64}$ .