jmc

The distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore, For $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities. For $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(0,2)$, $(2,0)$, $(1,3)$, and $(3,1)$, $8$ possibilities. For $z$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(0,2)$, $(0,4)$, $(2,0)$, $(4,0)$, $(2,4)$, $(4,2)$, $(1,3)$, and $(3,1)$, $13$ possibilities. However, we have $3\cdot 4\cdot 5=60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\frac{5\cdot 8\cdot 13-60}{60\cdot 59}=\frac{23}{177}⟹m+n=\boxed{200}$.