Let $\lfloor \sqrt{n} \rfloor = k$. Then $k^2 \leq n < (k+1)^2$. The only multiples of $k$ in the interval are $k^2, k(k+1), \text{ and } k(k+2)$. Thus $n$ must be in one of the aforementioned forms. It is can be shown that all of the forms work by reversing the steps. Thus the answer is all $n$ such that $n= k^2, k(k+1), k(k+2) \text{ for } k \in \mathbb{Z}^{+}$.

hood09 wrote: find all the integer $n\geq1$ such that $\lfloor\sqrt{n}\rfloor \mid n$ Make the replace $n=m^2+l$ where $0 \le l \le 2m$ then $\lfloor \sqrt{n} \rfloor=m$ so the divisibility condition becomes $m \mid m^2+l$ which is $m \mid l$ so $l=0,m,2m$ hence the answer set is $\boxed{n=m^2, m^2+m, m^2+2m}$ thus we are done