a) Let $n$ denote the number of red points. Then the following inequality holds (see (*) in the solution of Problem 2, Category C)
$$10 \cdot 21 - 4 \cdot 2 \le n \cdot 20 \le 10 \cdot 21 + 4 \cdot 2 \iff 202 \le 20n \le 218.$$
It is easy to see that there are no integers $n$ satisfying this condition.

b) We show that it is possible to mark $4$ blue, $10$ green, and $12$ red points on a plane so that the problem conditions hold.
For the convenience we multiply all distances by $4 \cdot 10 \cdot 12 = 480$. Let $A, B, C$ denote respectively the sums of the distances between red and green points, between blue and red points, green and blue points.
We have to construct the example with $A = 2 \cdot 480$, $B = 21 \cdot 480$, $C = 18 \cdot 480$. All points we place on the number axes.

First, we place $4$ blue points at the point $0$, $10$ green points at the point $216$, $6$ red points at the point $202$, and the other $6$ red points at the point $218$. Then $C = 4 \cdot 10 \cdot 216 = 18 \cdot 480$, $B = 4 \cdot 6 \cdot 202 + 4 \cdot 6 \cdot 218 = 21 \cdot 480$, and $A = 10 \cdot 6 \cdot (216 - 202) + 10 \cdot 6 \cdot (218 - 216) = 2 \cdot 480$.
It remains to move slightly the points so that all points become distinct but $A, B, C$ keep their values. We can replace $4$ blue points with the common coordinate $0$ by the points $-2\epsilon$, $-\epsilon$, $\epsilon$, $2\epsilon$; $10$ green points with the common coordinate $216$ by the points
$$216 - 5\epsilon, 216 - 4\epsilon, \dots, 216 - \epsilon, 216 + \epsilon, \dots, 216 + 5\epsilon;$$
$6$ red points with the common coordinate $202$ by the points
$$202 - 3\epsilon, 202 - 2\epsilon, 202 - \epsilon, 202 + \epsilon, 202 + 2\epsilon, 202 + 3\epsilon;$$
and, finally, $6$ red points with the common coordinate $218$ by the points
$$218 - 3\epsilon, 218 - 2\epsilon, 218 - \epsilon, 218 + \epsilon, 218 + 2\epsilon, 218 + 3\epsilon.$$
We can choose $\epsilon$ sufficiently small to keep the order of the points on the number axes.