Example 36 (6th CMO Question) Find all positive integers $n$, such that $\min _{k \in \mathbb{Z}^+}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$.

 untranslated part:
(No additional text to translate)
From $\alpha-1<[\alpha] \leqslant \alpha$, we know that $\min _{k \in \mathbb{N}_{+}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$ is equivalent to, for any $k \in \mathbf{N}_{+}$,
$$
k^{2}+\frac{n}{k^{2}} \geqslant k+\left[\frac{n}{k^{2}}\right] \geqslant 1991 .
$$

And there exists $k_{0} \in \mathbf{N}_{+}$, such that $k_{0}^{2}+\frac{n}{k_{0}^{2}}<1992$.
$$
\begin{array}{l}
\text { From (1) we have } k^{4}-1991 k^{2}+n \geqslant 0, \\
\left(k^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0, \\
n \geqslant \frac{1991^{2}}{4}-\left(k^{2}-\frac{1991}{2}\right)^{2} .
\end{array}
$$

To make this inequality hold for any $k \in \mathbf{N}_{+}$, $n$ must be no less than the maximum value of $\frac{1991^{2}}{4}-\left(k^{2}-\frac{1991}{2}\right)^{2}$. Since $32^{2}-\frac{1991}{2}=\frac{57}{2}<\frac{77}{2}=\frac{1991}{2}-31^{2}$, the maximum value of $\frac{1991^{2}}{4}-\left(k^{2}-\frac{1991}{2}\right)^{2}$ is $\frac{1991^{2}}{4}-\left(32^{2}-\frac{1991}{2}\right)^{2}=\frac{1991^{2}}{4}-\frac{57^{2}}{4}=967 \cdot 1024$, so $n \geqslant 967 \cdot 1024$.
From (2) we know that there exists $k_{0} \in \mathbf{N}_{+}$, such that $k_{0}^{4}-1992 k_{0}^{2}+n<0$,
$$
n<996^{2}-\left(k_{0}^{2}-996\right)^{2} \text {. }
$$

And the maximum value of $996^{2}-\left(k_{0}^{2}-996\right)^{2}$ is $996^{2}-\left(32^{2}-996\right)^{2}=968 \cdot 1024$, so $n<968 \cdot 1024$.

Therefore, $967 \cdot 1024 \leqslant n<968 \cdot 1024$, i.e., the set of values for $n$ is $\{n \mid 967 \cdot 1024 \leqslant n<968 \cdot 1024, n \in \mathbf{N}_{+}\}$.