Let the square be $A B C D$. First, suppose that all four vertices are closer to $X$ than $Y$. Then, by the triangle inequality, the sum of the scores is $A X+B X+C X+D X \geq A B+C D=2$. Similarly, suppose exactly two vertices are closer to $X$ than $Y$. Here, we have two distinct cases: the vertices closer to $X$ are either adjacent or opposite. Again, by the Triangle Inequality, it follows that the sum of the scores of the vertices is at least 2 . On the other hand, suppose that $A$ is closer to $X$ and $B, C, D$ are closer to $Y$. We wish to compute the minimum value of $A X+B Y+C Y+D Y$, but note that we can make $X=A$ to simply minimize $B Y+C Y+D Y$. We now want $Y$ to be the Fermat point of triangle $B C D$, so that \measuredangle B Y C=$\measuredangle C Y D=\measuredangle D Y B=120^{\circ}$. Note that by symmetry, we must have \measuredangle B C Y=\measuredangle D C Y=45^{\circ}$, so \measuredangle C B Y=\measuredangle C D Y=15^{\circ}$ And now we use the law of sines: $B Y=D Y=\frac{\sin 45^{\circ}}{\sin 120^{\circ}}$ and $C Y=\frac{\sin 15^{\circ}}{\sin 120^{\circ}}$. Now, we have $B Y+C Y+$ $D Y=\frac{\sqrt{2}+\sqrt{6}}{2}$, which is less than 2 , so this is our answer.