Squared away

The distance is optimized if every point lies on a corner of the square.

Let $A, B, C, D$ be the vertices of the square in clockwise order, and let $a, b, c, d$ be the number of points at each of these vertices. Then $D = 100\sqrt 2(ac + bd) + 100(ab + bc + cd + da).$ Claim: The distance is maximal if $a = b = c = d$ .

Proof: Treat $a, b, c, d$ as real numbers. Consider moving a small amount $dx$ from $A$ to $C$ , so that $da = -dx, dc = +dx, db = dd = 0.$ Then $dD = 100\sqrt 2(a\ dc + c\ da) + 100(b\ da + b\ dc + d\ dc + d\ da = 100\sqrt2(a -c)dx.$ If the situation is maximal, this value must be zero, so that $a = c$ With a similar reasoning we find $b = d$ . We rewrite the situation as $dD = 100\sqrt 2(a^2 + b^2) + 400ab.$

Now consider moving a small amount $dy$ from $A/C$ to $B/D$ , so that $da = -dy, db = +dy$ . Then $dD = 100\sqrt 2(2a\ da + 2b\ db) + 400(a\ db + b\ da) = (200\sqrt 2-400)(b - a)dy.$ Again, this must be zero in the maximal situation so that $a = b$ .

This proves that $a = b = c = d$ , and since $a + b + c + d = 12$ this means that each vertex of the square contains three of the points.

Finally, since $ab = ac = ad = bc = bd = cd = 3\cdot 3 = 9,$ $D = 100\sqrt 2(9 + 9) + 100(9 + 9 + 9 + 9) \approx \boxed{6145}.584412.$