On a past  Mathematical Olympiad  the maximum  possible  score  on a  problem was 5. The  average score of boys  was 4,  the average  score  of girls  was 3.25, and the overall  average  score  was  3.60. Find the total number  of participants, knowing  that  it was  in the range  from  31  to 50.
1. Let \( b \) be the number of boys and \( g \) be the number of girls.
2. The average score of boys is 4, so the total score of boys is \( 4b \).
3. The average score of girls is 3.25, so the total score of girls is \( 3.25g \).
4. The overall average score is 3.60, so the total score of all participants is \( 3.6(b + g) \).
5. We set up the equation for the total scores:
   \[
   4b + 3.25g = 3.6(b + g)
   \]
6. Distribute and simplify the equation:
   \[
   4b + 3.25g = 3.6b + 3.6g
   \]
   \[
   4b - 3.6b = 3.6g - 3.25g
   \]
   \[
   0.4b = 0.35g
   \]
7. Solve for \( b \) in terms of \( g \):
   \[
   b = \frac{0.35}{0.4}g = \frac{7}{8}g
   \]
8. Since \( b \) and \( g \) must be integers, \( g \) must be a multiple of 8.
9. The total number of participants \( b + g \) must be in the range from 31 to 50.
10. Let \( g = 8k \) for some integer \( k \). Then \( b = \frac{7}{8}g = 7k \).
11. The total number of participants is:
    \[
    b + g = 7k + 8k = 15k
    \]
12. We need \( 31 \leq 15k \leq 50 \). Solving for \( k \):
    \[
    31 \leq 15k \leq 50
    \]
    \[
    \frac{31}{15} \leq k \leq \frac{50}{15}
    \]
    \[
    2.0667 \leq k \leq 3.3333
    \]
13. Since \( k \) must be an integer, the only possible value for \( k \) is 3.
14. Therefore, \( g = 8 \times 3 = 24 \) and \( b = 7 \times 3 = 21 \).
15. The total number of participants is:
    \[
    b + g = 21 + 24 = 45
    \]

The final answer is \(\boxed{45}\).