Satisfy $m n \geqslant 0, m^{3}+n^{3}+99 m n=33^{3}$ the historical integer pairs $(m, n)$ are $\qquad$ pairs.
12. 35

If $m+n=s$, then
$$
\begin{array}{c}
m^{3}+n^{3}+3 m n(m+n)=s^{3} \\
m^{3}+n^{3}+99 m n=33^{3}
\end{array}
$$

Subtracting (2) from (1) gives $s^{4}-33^{3}=3 m n s-99 m n$, which means $(s-33)\left(s^{2}+33 s+3 s^{2}-3 m n\right)=0$, so $s=$ 33 or $s^{2}+33 s-33^{2}-3 m n=0$.

If $m+n=33$ and $m n \geqslant 0$, there are 34 solutions: $(m, n)=(0,33),(1,32), \cdots,(33,0)$. From $s^{2}$ $-33 s+33^{2}-3 m n=0$, we get $(m+n)^{2}+33(m-n)+33^{2}-3 m n=0,(m-n)^{2}+(m+33)^{2}+$ $(n+33)^{2}=0$, which has the unique solution $(-33,-33)$. In total, there are 35 solutions.