Example 6 (CMO-6 Test) Find all natural numbers $n$ such that
$$
\min _{k \in \mathbf{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991(n \in \mathbf{N}) .
$$
The given condition is equivalent to: for all $k \in \mathbf{N}$,
$$
k^{2}+\left(n / k^{2}\right) \geqslant 1991 \text {, }
$$

and there exists $k \in \mathbf{N}$, such that $k^{2}+\left(n / k^{2}\right)<1992$.
(1) is equivalent to for all $k \in \mathbf{N}$,
$k^{4}-1991 k^{2}+n \geqslant 0$, which is $\left(k^{2}-1991 / 2\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0$.
Since $[\sqrt{1991 / 2}]=31$, and $31^{2}-\frac{1991}{2}=-\frac{69}{2}, 32^{2}-\frac{1991}{2}=\frac{57}{2}$,
the left side of the above inequality is minimized at $k=32$, so (1) is equivalent to
$$
n \geqslant 1991 \cdot 32^{2}-32^{4}=1024 \cdot 967 \text {. }
$$

(2) is equivalent to there exists $k \in \mathbf{N}$, such that $\left(k^{2}-996\right)^{2}+n-996^{2}<0$.
The left side of the above inequality is also minimized at $k=32$, so (2) is equivalent to
$$
n<1992 \cdot 32^{2}-32^{4}=1024 \cdot 968 \text {. }
$$

Therefore, all the required $n$ are: $1024 \cdot 967 \leqslant n \leqslant 1024 \cdot 967+1023(n \in \mathbf{N})$.