jmc

Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$. Then $\angle AED=180-53-37=90^{\circ }$. As $\angle AED=90^{\circ }$, note that the midpoint of $\overline{AD}$, $N$, is the center of the circumcircle of $△AED$. We can do the same with the circumcircle about $△BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that $$NE=ND=\frac{AD}{2}=1004,ME=MC=\frac{BC}{2}=500.$$ Thus $MN=NE-ME=\boxed{504}$. For purposes of rigor we will show that $E,M,N$ are collinear. Since $\overline{BC}\parallel \overline{AD}$, then $BC$ and $AD$ are homothetic with respect to point $E$ by a ratio of $\frac{BC}{AD}=\frac{125}{251}$. Since the homothety carries the midpoint of $\overline{BC}$, $M$, to the midpoint of $\overline{AD}$, which is $N$, then $E,M,N$ are collinear.