14.59 Find all natural numbers $n$ such that $\min _{k \in N}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=1991$, where $N$ is the set of natural numbers.
(6th China High School Mathematics Winter Camp, 1991)
[Solution] Since
$$
k^{2}+\frac{n}{k^{2}}-10, \\
n<-\left(k_{0}^{2}-996\right)^{2}+996^{2} .
\end{array}
$$

That is,

to make this inequality hold, we only need to find the value of $k_{0}$ that maximizes $-\left(k_{0}^{2}-996\right)^{2}+996^{2}$ and is greater than $n$.

Clearly, when $k_{0}=32$, $-\left(k_{0}^{2}-996\right)^{2}+996^{2}$ reaches its maximum value $-(1024-996)^{2}+996^{2}$, which is
$$
n<-(1024-996)^{2}+996^{2}=1024 \cdot 968 .
$$

Thus, $\quad 1024 \cdot 967 \leqslant n<1024 \cdot 968$.
Therefore, the set $\{n \mid 1024 \cdot 967 \leqslant n<1024 \cdot 968, n \in N\}$ is the solution.