Solution:

Since all vertices are lattice points, $w = \overrightarrow{BC}$ is an integer vector with $|w| = |\overrightarrow{AD}| = \sqrt{50},$ so $w$ is one of $(\pm 1, \pm 7),$ $(\pm 7, \pm 1),$ $(\pm 5, \pm 5).$ Write $\overrightarrow{AD} = (1, 7) = s\,\hat{u} + h\,\hat{v}$ where $\hat{u}$ points along $\overline{AB}$ and $\hat{v}$ is perpendicular. Because $\overline{AB} \parallel \overline{CD},$ the vector $w$ has the same perpendicular component $h,$ and the equal leg lengths force its $\hat{u}$-component to be $-s$ (the value $+s$ gives a parallelogram). Hence $(1, 7) - w = 2s\,\hat{u}$ is parallel to $\overline{AB}.$

Discard $w = (1, 7)$ (parallelogram) and $w = (-1, -7)$ (then $h = 0,$ degenerate). The choices $w = (1, -7)$ and $w = (-1, 7)$ make $(1, 7) - w$ vertical or horizontal, which is forbidden. The remaining eight choices give $(1, 7) - w$ equal to $(-6, 6),$ $(-6, 8),$ $(8, 6),$ $(8, 8),$ $(-4, 2),$ $(-4, 12),$ $(6, 2),$ $(6, 12),$ with slopes $-1,$ $-\frac{4}{3},$ $\frac{3}{4},$ $1,$ $-\frac{1}{2},$ $-3,$ $\frac{1}{3},$ $2;$ each is realizable by placing $B$ suitably far along $\hat{u}.$

The sum of the absolute values is $1 + \frac{4}{3} + \frac{3}{4} + 1 + \frac{1}{2} + 3 + \frac{1}{3} + 2 = \frac{119}{12},$ so $m + n = 119 + 12 = 131.$