12. 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 $\mathbb{N}$ is the set of natural numbers.
12. Since 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 $k^{2}+\frac{n}{k^{2}} \geqslant 1991$, thus
$$
k^{4}-1991 k^{2}+n \geqslant 0,\left(k^{2}-\frac{1991}{2}\right)^{2}+n-\frac{1991^{2}}{4} \geqslant 0 .
$$

When $k=32$, $k^{2}-\frac{1991}{2}=\frac{57}{2},\left(k^{2}-\frac{1991}{2}\right)^{2}$ reaches the minimum value $\frac{57^{2}}{4}$, hence,
$$
n \geqslant \frac{1}{4}\left(1991^{2}-57^{2}\right)=\frac{1}{4}(1991+57)(1991-57)=1024 \cdot 967 \text {. }
$$

On the other hand, the minimum value of $k^{2}+\left[\frac{n}{k^{2}}\right]$ is 1991, which means there exists a natural number $k$ such that $k^{2}+\frac{n}{k^{2}}$ $<1992$, i.e., $\left(k^{2}-996\right)^{2}+n-996^{2}<0$. When $k=32$, $\left(k^{2}-996\right)^{2}=28^{2}$ reaches the minimum value, then we should have $n<996^{2}-28^{2}=1024 \cdot 968$, so the natural number $n$ satisfies $1024 \times 967 \leqslant n$ $\leqslant 1024 \times 967+1023$