还是把我的解交上来供大家审阅吧

Fact1. If N >= 4, there is a pair of A and B, such that AB <> 1.

Fact2. Pick up AB <> 1, Draw unit circle A and B (centered at A and B). The rest points must be on one of the circles.

Fact3. If all points are on a unit circle, then maximal N is 4. Proof is tedious. You can drawing and trying.

Given unit Equilateral triangle ABC. Draw unit Equilateral triangle on AB, BC, CA, get D, E, F. Call D, E, F  associated point of unit Equilateral triangle ABC.

Fact4. Given a unit circle O and a unit Equilateral triangle ABC. If none of A, B, C is on the circle O, then there is at most one associated point of ABC is no circle O, because the distance between the associated points is 2.

Now prove N=8 is impossible.

Suppose AB<>1. Draw unit Circle A and B as in fact2. Consider circle A. Look at the rest 6 points. It is from fact3 that there are only two cases.

Case 1: 4 points C, D, E, F on circle A, and 2 points G, H not on circle A. Apply fact 2 to AB, AG and AH, BGH form a unit Equilateral triangle. From 4 points on  circle A, one can find, say, CE, DF ( or CF, DE ...), such that both of them have length not 1. Consider BCE, GCE and HCE, two of B, G, H, must have distance 1 to one of C, E. In other word, one of C, E is the associated point of BGH. The same is to DF. Then there are two associated points of BGH on circle A. It is a contradiction to fact 4.

Case 2: 3 points C, D, E on circle A, and 3 points F, G, H not on circle A. It is impossible because of fact 2.

Note: In the construction of 7, two basic unit circles are A and D. The rest 5 points, 4 are on circle A, and 1 is on D. If add one more on A, it is a contradiction to fact 2. If add one on D, this is our case 1. If move one point from A to D, you cannot even construct 7, from  康MM guess (and I believe it).