Example 25 (1991 China Mathematical Olympiad) Find all natural numbers $n$ such that
$$
\min _{k \in \mathbb{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991,
$$

where $\left[\frac{n}{k^{2}}\right]$ denotes the greatest integer not exceeding $\frac{n}{k^{2}}$, and $\mathbf{N}$ is the set of natural numbers.
The condition $\min _{k \in \mathrm{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$ is equivalent to the following two conditions:
(1) For any $k \in \mathbf{N}$, we have $k^{2}+\frac{n}{k^{2}} \geqslant 1991$;
(2) There exists $k_{0} \in \mathbf{N}$, such that $k_{0}^{2}+\frac{n}{k_{0}^{2}}<1992$.

Conditions (1) and (2) are also respectively equivalent to:
(1) For any $k \in \mathbf{N}$, we have $k^{4}-1991 k^{2}+n \geqslant 0$;
(2) There exists $k_{0} \in \mathbf{N}$, such that $k_{0}^{4}-1992 k_{0}^{2}+n<0$.
We first solve the inequality in (1), which gives
$$
\left(k^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0 \text {, }
$$

Since the closest perfect square to $\frac{1991}{2}=995.5$ is $32^{2}=1024$, from (1) we have
$$
\left(1024-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0 \text {. }
$$

Solving this, we get $n \geqslant 1024 \times 967=990208$.
Condition (2) is equivalent to
$$
\min _{k \in \mathrm{N}}\left\{\left(k^{2}-996\right)^{2}+n-996^{2}\right\}<0,
$$

Since the minimum value of $\left(k^{2}-996\right)^{2}$ is $28^{2}$, from (2) we get
$$
n<996^{2}-28^{2}=1024 \times 968=991232 \text {. }
$$

In summary, the range of $n$ that satisfies the problem's requirements is $990208 \leqslant n \leqslant 991231$.