$2 \cdot 94$ Find all natural numbers $n$ such that
$$\min _{k \in N}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$$

where $\left[\frac{n}{k^{2}}\right]$ denotes the greatest integer not exceeding $\frac{n}{k^{2}}$, and $N$ is the set of natural numbers.
[Solution] The condition $\min _{k \in N}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$ is equivalent to the following two conditions:
(1) For any $k \in N$, $k^{2}+\frac{n}{k^{2}} \geqslant 1991$;
(2) There exists $k_{0} \in N$ such that $k_{0}^{2}+\frac{n}{k_{0}^{2}}<1992$. Conditions (1) and (2) are also respectively equivalent to
(1') For any $k \in N$, $k^{4}-1991 k^{2}+n \geqslant 0$;
(2') There exists $k_{0} \in N$ such that $k_{0}^{4}-1992 k_{0}^{2}+n<0$.
We first solve the inequality in (1'), at this time we have
$$\left(k^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0$$

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

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

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

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