250. Let the line $M N$ intersect $D_{1} C_{1}$ at point $L$. Denote: $|A M|=x,|B N|=y$. From the condition, it follows that $x>a, y>a$. By projecting all points onto the plane $A B B_{1} A_{1}$, we find $\frac{\left|C_{1} L\right|}{\left|L D_{1}\right|}=$
$=\frac{a}{x-a}$, and by projecting them onto the plane $A B C D$, we find $\frac{\left|C_{1} L\right|}{\left|L D_{1}\right|}=\frac{y-a}{a}$. Therefore, $\frac{a}{x-a}=\frac{y-a}{a}$, from which $x y=$ $=(x+y) a$. But $(x+y)^{2} \geqslant 4 x y$. This means, $x y \geqslant 4 a^{2}$.

Now we obtain $|M N|^{2}=x^{2}+y^{2}+a^{2}=(x+y)^{2}-2 x y+a^{2}=$ $=\frac{(x y)^{2}}{a^{2}}-2 x y+a^{2}=\frac{1}{a^{2}}\left(x y-a^{2}\right)^{2} \geqslant 9 a^{2}$. The smallest value of $|M N|$ is $3 a$.
250. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge $a$. On the line $A A_{1}$, a point $M$ is taken, and on the line $B C$ - a point $N$ such that the line $M N$ intersects the edge $C_{1} D_{1}$. Find the minimum value of the magnitude | $M N \mid$.