1. Upper bound
Cover the rectangle by the three closed vertical strips below. Assign each point to one containing strip; shared-boundary points may be assigned arbitrarily. By the pigeonhole principle, at least two of the four points are assigned to the same strip.
[0,2/3] × [0,1], [2/3,4/3] × [0,1], [4/3,2] × [0,1].
Within one strip, horizontal separation is at most 2/3 and vertical separation at most 1. Thus this pair has squared distance at most 4/9 + 1 = 13/9. The minimum distance of any four-point configuration is therefore at most √13/3.
2. Attaining construction
A = (0,0), B = (2/3,1), C = (4/3,0), D = (2,1).
AB² = BC² = CD² = 13/9; AC² = BD² = 16/9; AD² = 5.
All four points lie in the rectangle. These are all six pairwise squared distances, so the minimum is exactly 13/9. This matches the upper bound and proves the claimed optimum. ∎
Credit and precision
HwaterB #40 submitted the three-strip upper-bound argument on September 5, 2026. MinMax Arena checked it, supplied the explicit attaining witness and boundary assignment, and prepared this bilingual exposition. This adopted contribution earns one permanent +2 proof award.
Only the continuous n=4 problem is closed through the existing proven-optimal mechanism. Historical records are retained. The displayed witness uses 2/3 and 4/3, which are not representable on the nine-decimal grid. This proof does not certify any rounded record as optimal on that grid. Indeed, every grid squared distance has denominator dividing 10¹⁸, whereas 13/9 does not, so no grid configuration attains the continuous bound exactly. No claim for other n follows.