[Example 5.3.6] Find all natural numbers $n$ such that
$$
\min _{k \in \mathbf{N}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991 \text {, }
$$

where $\left[\frac{n}{k^{2}}\right]$ denotes the greatest integer not exceeding $\frac{n}{k^{2}}$, and $\mathbf{N}$ is the set of positive integers.
How to understand equation (1)? We treat $k$ in (1) as the variable $x$, and let
$$
f(x)=x^{2}+\frac{n}{x^{2}} \text {. }
$$

Since $\left[\frac{n}{x^{2}}\right] \leqslant \frac{n}{x^{2}}$, then (1) is equivalent to
for any positive integer $x$, we have
$$
f(x) \geqslant 1991,
$$

and there exists a positive integer $x_{0}$ such that
$$
f\left(x_{0}\right)<1992 .
$$

Substituting $f(x)$, we get the more specific inequalities:
for any positive integer $x$, we have
$$
x^{4}-1991 x^{2}+n \geqslant 0,
$$

there exists a positive integer $x_{0}$ such that
$$
x_{0}^{4}-1992 x_{0}^{2}+n<0 .
$$

The condition for (2) to always hold is that the minimum value of $x^{4}-1991 x^{2}+n$ is greater than or equal to 0, and the condition for (3) to hold is that the minimum value of $x^{4}-1992 x^{2}+n$ is less than 0.
Thus, the problem is transformed into finding the range of $n$ such that
$$
\left\{\begin{array}{l}
\min _{x \in \mathbf{N}}\left(x^{4}-1991 x^{2}+n\right) \geqslant 0, \\
\min _{x \in \mathbb{N}}\left(x^{4}-1992 x^{2}+n\right)<0 .
\end{array}\right.
$$

Next, we find the minimum values of $x^{4}-1991 x^{2}+n$ and $x^{4}-1992 x^{2}+n$.
$$
\begin{array}{r}
x^{4}-1991 x^{2}+n=\left(x^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4}, \\
x^{4}-1992 x^{2}+n=\left(x^{2}-996\right)^{2}+n-996^{2} .
\end{array}
$$

Noting that $\frac{1991}{2}$ and 996 are closest to the square number $32^{2}=1024$, the minimum values of the above two expressions are
$$
\begin{aligned}
\min _{x \in \mathbf{N}}\left(x^{4}-1991 x^{2}+n\right) & =\left(32^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \\
& =n-967 \times 1024 . \\
\min _{x \in \mathrm{N}}\left(x^{4}-1992 x^{2}+n\right) & =\left(32^{2}-996\right)^{2}+n-996^{2} \\
& =n-968 \times 1024 .
\end{aligned}
$$

Thus, we have
$$
\begin{array}{c}
\left\{\begin{array}{l}
n-967 \times 1024 \geqslant 0 \\
n-968 \times 1024<0
\end{array}\right. \\
967 \times 1024 \leqslant n<968 \times 1024 .
\end{array}
$$

This is the range of natural numbers $n$ required by the problem.
In solving this problem, understanding the given equation is key, and the transformation of the proposition through the understanding of (1) is crucial.