Belarusian Mathematical Olympiad

Let $n$, $k$, $m$ denote respectively the number of red, green, blue points. Then $m + n + k = 15$. From (*) in the solution of Problem 2, Category C, we have
$$39m \le 51k + n, \quad (1)$$
$$51k - n \le 39m. \quad (2)$$
Substitute for $k$ in (1) to obtain (after simple manipulations)
$$18m + 10n \le 153 \Rightarrow 9m + 5n \le 76. \quad (3)$$
Substitute for $k$ in (2) to obtain $51 \cdot 15 \le 90m + 52n$. Using $n \le 13$, we obtain $51 \cdot 15 \le 90m + 50n + 26$, hence
$$74 \le 9m + 5n. \quad (4)$$
If $n \le 7$, then we have $51 \cdot 15 \le 90m + 50n + 14$ or $76 \le 9m + 5n$. From (3), it follows that $9m + 5n = 76$. However, it is easy to verify that the last equality cannot be valid for $n \le 7$.
Verifying the values $n = 8, 9, \dots, 13$, we see that inequalities (3) and (4) can be valid only if $n = 8$ or $n = 13$. Then we have respectively $m = 4$, $k = 3$ and $m = k = 1$. Examples for these values can be constructed similarly to the example of Problem 2, Category B.

Therefore, the answer is $(n, k, m) = (8, 3, 4)$ or $(13, 1, 1)$.