Find all natural numbers $n$ such that
\[
\min _{k \in \mathbb{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991
\]

Here $\left[\frac{n}{k^{2}}\right]$ denotes the greatest integer not exceeding $\frac{n}{k^{2}}$, and $\mathbf{N}$ is the set of natural numbers.
Given that when $k$ takes all natural numbers, the minimum value of $k^{2}+\left[\frac{n}{k^{2}}\right]$ is 1991, for all natural numbers $k$ we have
$$
k^{2}+\frac{n}{k^{2}} \geqslant 1991
$$

Therefore,
$$
\begin{array}{c}
k^{4}-1991 k^{2}+n \geqslant 0 \\
\left(k^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0
\end{array}
$$

When $k=32$, $k^{2}-\frac{1991}{2}=\frac{57}{2}$ makes $\left(k^{2}-\frac{1991}{2}\right)^{2}$ reach a small minimum value of $\frac{57^{2}}{4}$. Therefore, we should have
$$
\begin{aligned}
n \geqslant & \frac{1}{4}\left(1991^{2}-57^{2}\right)= \\
& \frac{1}{4} \cdot(1991+57)(1991-57)=1024 \cdot 967
\end{aligned}
$$

On the other hand, the minimum value of $k^{2}+\left[\frac{n}{k^{2}}\right]$ being 1991 also means there exists a natural number $k$ such that

i.e.,
$$
\begin{array}{c}
k^{2}+\frac{n}{k^{2}}<1992 \\
k^{4}-1992 k^{2}+n<0 \\
\left(k^{2}-966\right)^{2}+n-996^{2}<0
\end{array}
$$

or
$$
\begin{array}{c}
k^{2}+\frac{n}{k^{2}}<1992 \\
k^{4}-1992 k^{2}+n<0 \\
\left(k^{2}-966\right)^{2}+n-996^{2}<0
\end{array}
$$

When $k=32$, $\left(k^{2}-996\right)^{2}=28^{2}$ reaches its minimum value, so we should have $n \leqslant 996^{2}-$ $28^{2}=1024 \cdot 968, \cdots$ Therefore, the natural number $n$ satisfies
$$
1024 \cdot 967 \leqslant n \leqslant 1024 \cdot 967+1023
$$