More generally, suppose three vertices of the square lie on lines $y=y_{1}, y=y_{2}, y=y_{3}$. One of these vertices must be adjacent to two others. If that vertex is on $y=y_{1}$ and the other two are on $y=y_{2}$ and $y=y_{3}$, then we can use the Pythagorean theorem to get that the square of the side length is $(y_{2}-y_{1})^{2}+(y_{3}-y_{1})^{2}$. For $(y_{1}, y_{2}, y_{3})=(2,0,18)$, the possibilities are $2^{2}+16^{2}, 2^{2}+18^{2}, 16^{2}+18^{2}$, so the sum is $2(2^{2}+16^{2}+18^{2})=2(4+256+324)=1168$.