14. Find all positive integers $n$ such that $\min _{k \in \mathbb{N}_{+}}\left(k^{2}+\left[\frac{n}{k^{2}}\right]\right)=2006$.
14. Solution: The condition in the problem $\Leftrightarrow$ for all $n \in \mathbf{N}_{+}$, we have $k^{2}+\frac{n}{k^{2}} \geqslant 2006$

and there exists $k \in \mathbf{N}_{+}$ such that $k^{2}+\frac{n}{k^{2}}<2007$
(1) $\Leftrightarrow$ for all $k \in \mathbf{N}_{+}$, we have $k^{4}-2006 k^{2}+n \geqslant 0$,

which is $\left(k^{2}-1003\right)^{2}+n-1003^{2} \geqslant 0$
Since $[\sqrt{1003}]=31$, and $31^{2}-1003=-42, 32^{2}-1003=21$.
Thus, the left side of (3) is minimized when $k=32$, so (1) $\Leftrightarrow n \geqslant 2006 \times 32^{2}-32^{4}=1024 \times 982$.
Also, (2) $\Leftrightarrow$ there exists $k \in \mathbf{N}_{+}$ such that $\left(k^{2}-\frac{2007}{2}\right)^{2}+n-\left(\frac{2007}{2}\right)^{2}<0$,
the left side of the above equation is also minimized when $k=32$, so (2) $\Leftrightarrow n<2007 \times 32^{2}-32^{4}=1024 \times 983$.
Therefore, $n$ is any integer satisfying $1024 \times 982 \leqslant n \leqslant 1024 \times 982+1023$.