11. If $n \in \mathbf{N}^{+}$, and $[\sqrt{n}] \mid n$, then such $n$ is
11. $n=k^{2}, k^{2}+k, k^{2}+2 k\left(k \in \mathbf{N}^{*}\right)$. Hint: We can set $n=k^{2}+r, 0 \leqslant r \leqslant 2 k$, then $k \mid r$, so, $r=0$ or $k$, or $2 k$.