2008 AIME II Problem 5

Extend the legs $AB$ and $DC$ to meet at $E;$ since $37^\circ + 53^\circ = 90^\circ,$ the angle at $E$ is right

The median to the hypotenuse of a right triangle is half the hypotenuse, and $E,$ $M,$ $N$ are collinear

Extend legs $\overline{AB}$ and $\overline{DC}$ until they meet at a point $E.$ Since $\angle A + \angle D = 37^\circ + 53^\circ = 90^\circ,$ triangle $EAD$ has a right angle at $E.$ Because $\overline{BC} \parallel \overline{AD},$ triangle $EBC$ is the image of triangle $EAD$ under a homothety centered at $E,$ so the midpoint $M$ of $\overline{BC}$ maps to the midpoint $N$ of $\overline{AD};$ in particular $E,$ $M,$ and $N$ are collinear.

The median to the hypotenuse of a right triangle is half the hypotenuse, so $EN = \frac{2008}{2} = 1004$ and $EM = \frac{1000}{2} = 500.$ Therefore $$\begin{aligned} MN &= EN - EM \\ &= 1004 - 500 = 504. \end{aligned}$$