If we take any 4 blue points (convex or non-convex) there exist at most 2 good triangles. Then it's easy to find max number of good triangles and number of all triangles with vertices at blue points. Then we can show that number of max good triangles = half of all traingles with vertices at blue points. But we need to show that there isn't equality case. Which means we should show that any 4 blue points can't make 2 good triangles.
Take a line through red point. Then at most 1 blue point can situated on this line. Then there exist at least 49 points which situated same side of line. Then take 4 of them. These 4 points can't make any good triangle. Then, there isn't equality case, which means number of good triangles is less than the half of the total number of triangles with vertices at blue points.