jmc

For simplicity, we translate the points so that $A$ is on the origin and $D=(1,7)$. Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$, and let $D^{\prime }=(a,b)$ be the reflection of $D$ across that perpendicular. Then $ABC{D}^{\prime }$ is a parallelogram, and $\overrightarrow{AB}=\overrightarrow{D^{\prime }C}$. Thus, for $C$ to have integer coordinates, it suffices to let $D^{\prime }$ have integer coordinates.[1] Let the slope of the perpendicular be $m$. Then the midpoint of $\overline{D{D}^{\prime }}$ lies on the line $y=mx$, so $\frac{b+7}{2}=m\cdot \frac{a+1}{2}$. Also, $AD=A{D}^{\prime }$ implies that $a^2+b^2=1^2+7^2=50$. Combining these two equations yields $$a^2+{\left(7-(a+1)m\right)}^2=50$$ Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $(\pm 1,\pm 7),(\pm 7,\pm 1),(\pm 5,\pm 5)$. We exclude the cases $(\pm 1,\pm 7)$ because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have $$7-8m=\pm 1,7+6m=\pm 1,7-6m=\pm 5,7+4m=\pm 5$$ These yield $m=1,\frac{3}{4},-1,-\frac{4}{3},2,\frac{1}{3},-3,-\frac{1}{2}$, and the sum of their absolute values is $\frac{119}{12}$. The answer is $m+n=\boxed{131}$.