12. The positive integer solutions $(x, y)$ of the equation $2 x^{2}-x y-3 x+y+2006=0$ are $\qquad$ pairs.


12.4.

From $2 x^{2}-x y-3 x+y+2006=0$, we get $y=\frac{2 x^{2}-3 x+2006}{x-1}=2 x-1+\frac{2005}{x-1}$.

Therefore, $x-1$ can take the values $1,5,401,2005$. Hence, the equation has 4 pairs of positive integer solutions.