Denote $f_k(n)=k^2+\lfloor \frac{n}{k^2}\rfloor$. Consider $g_k(n)=\frac{f_{k+1}(n)}{f_k(n)}$ in both cases $g_k(n)\ge 1$ and $g_{k-1}(n)\le 1$ we get \[ k^2(k+1)^2\ge n\ge k^2(k-1)^2 (*)\]. Let $i$ be positive integer satisfying (*), then $i^2+\lfloor \frac{n}{i^2}\rfloor = f_i(n) =\min_{k\in \mathbb{N}} f_k(n)$. From the given condition, $\min_{k\in \mathbb{N}} f_k(n)=1991$, we get: \[ \begin{cases} i^2(i-1)^2\le n\le i^2(i+1)^2 \\ (1991-i^2)i^2\le n\le i^2(1992-i^2)-1 \end{cases} (**)\] - If $i^2(i-1)^2\le (1991-i^2)i^2\Rightarrow i\le 32$. Thus for $i\le 31$, $ i^2(i+1)^2\le i^2(1992-i^2)-1$, then \[ (*)\Leftrightarrow (1991-i^2)i^2\le n\le i^2(i+1)^2\Rightarrow (1991-i^2)i^2\le i^2(i+1)^2, \forall, i\in \mathbb{N}, i\le 31\], contradiction. Thus $i=32$, and $ i^2(i+1)^2\ge i^2(1992-i^2)-1$. Hence, $990208\le n\le 991231$ is a solution. - If $i^2(i-1)^2\ge (1991-i^2)i^2$, similarly, we get contradiction and no solution. Finally, $n\in\mathbb{Z}, 990208\le n\le 991231$ is solution.