12.B. Given that $a$ and $b$ are positive integers, the quadratic equation $x^{2}-2 a x+b=0$ has two real roots $x_{1}$ and $x_{2}$, and the quadratic equation $y^{2}+2 a y+b=0$ has two real roots $y_{1}$ and $y_{2}$. It is also given that $x_{1} y_{1}-x_{2} y_{2}=2008$. Find the minimum value of $b$.
12.B. For the equation in $x$, $x^{2}-2 a x+b=0$, the roots are $a \pm \sqrt{a^{2}-b}$. For the equation in $y$, $y^{2}+2 a y+b=0$, the roots are $-a \pm \sqrt{a^{2}-b}$.
Let $\sqrt{a^{2}-b}=t$.
Then when $x_{1}=a+t, x_{2}=a-t, y_{1}=-a+t$, $y_{2}=-a-t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the condition;

When $x_{1}=a-t, x_{2}=a+t, y_{1}=-a-t$, $y_{2}=-a+t$, we have $x_{1} y_{1}-x_{2} y_{2}=0$, which does not satisfy the condition;

When $x_{1}=a-t, x_{2}=a+t, y_{1}=-a+t$, $y_{2}=-a-t$, we get $x_{1} y_{1}-x_{2} y_{2}=4 a t$;

When $x_{1}=a+t, x_{2}=a-t, y_{1}=-a-t$, $y_{2}=-a+t$, we get $x_{1} y_{1}-x_{2} y_{2}=-4 a t$.
Since $t=\sqrt{a^{2}-b}>0$, we have $a t=502$.
Given that $a$ is a positive integer, we know that $t$ is a rational number, and thus, $t$ is an integer.

From $a t=502$, we get $a=251, t=2$, so the minimum value of $b$ is $b=a^{2}-t^{2}=251^{2}-2^{2}=62997$.
Therefore, the minimum value of $b$ is 62997.