There are $$n$$ points in the plane, no three of which are collinear. Every pair of points is joined by a segment which is coloured red or blue such that the following conditions hold:
(a) If $$A, B, C$$ are three points such that $$AB$$ is red and $$BC$$ is blue, then $$AC$$ is red.
(b) For any point $$A$$, there are exactly three points $$B, C, D$$ such that $$AB, AC, AD$$ are red.
Find the sum of all possible values of $$n$$.

Take a point $$A$$ with red neighbours $$B, C, D$$. For any other point $$X$$ the segment $$AX$$ is blue, and applying the first condition to $$B, A, X$$ forces $$BX$$ to be red, so $$B$$ has at least $$1 + (n - 4) = n - 3$$ red neighbours and $$n \le 6$$, while clearly $$n \ge 4$$. Counting endpoints of red segments, $$3n$$ must be even, so $$n$$ is even, leaving $$n = 4$$ and $$n = 6$$. Both occur, by colouring every segment red when $$n = 4$$, and by splitting six points into two groups of three with blue inside groups and red between them, so the sum is $$4 + 6 = 10$$.