There are 100 boxers, each of them having different strengths, who participate in a tournament. Any of them fights each other only once. Several boxers form a plot. In one of their matches, they hide in their glove a horse shoe. If in a fight, only one of the boxers has a horse shoe hidden, he wins the fight; otherwise, the stronger boxer wins. It is known that there are three boxers who obtained (strictly) more wins than the strongest three boxers. What is the minimum number of plotters ?

[i]Proposed by N. Kalinin[/i]
1. **Define the problem and variables:**
   - Let \( p \) be the number of plotters.
   - Let \( u \) be the number of usual boxers (those who are neither the three strongest nor plotters).
   - The three strongest boxers are denoted as \( s_1, s_2, \) and \( s_3 \) in decreasing order of strength.
   - The plotters are denoted as \( s_4, s_5, \ldots, s_{p+3} \).

2. **Assumptions and initial conditions:**
   - The plotters should not use horseshoes against usual boxers since it does not affect the number of wins of the strongest boxers.
   - Each plotter is stronger than the usual boxers, so they win all their matches against them.
   - The three strongest boxers \( s_1, s_2, s_3 \) will have a certain number of wins among themselves:
     - \( s_1 \) wins against \( s_2 \) and \( s_3 \), so \( s_1 \) has 2 wins.
     - \( s_2 \) wins against \( s_3 \), so \( s_2 \) has 1 win.
     - \( s_3 \) has 0 wins.

3. **Wins calculation for plotters:**
   - Each plotter \( s_i \) (for \( i = 4, 5, \ldots, p+3 \)) will win against all usual boxers and other plotters weaker than themselves.
   - \( s_4 \) will have \( p-1 \) wins among the plotters.
   - \( s_5 \) will have \( p-2 \) wins among the plotters.
   - \( s_6 \) will have \( p-3 \) wins among the plotters.

4. **Using horseshoes:**
   - Each plotter \( s_4, s_5, s_6 \) should use a horseshoe against one of \( s_1, s_2, s_3 \) to maximize their wins.
   - Therefore, \( s_4 \) will have \( u + (p-1) + 1 \) wins.
   - \( s_5 \) will have \( u + (p-2) + 1 \) wins.
   - \( s_6 \) will have \( u + (p-3) + 1 \) wins.

5. **Minimizing the wins of \( s_1, s_2, s_3 \):**
   - To minimize the wins of \( s_1, s_2, s_3 \), plotters should use horseshoes against them.
   - Let \( \frac{p}{3} + 1 \) plotters use horseshoes against \( s_1 \).
   - Let \( \frac{p}{3} \) plotters use horseshoes against \( s_2 \).
   - Let \( \frac{p}{3} - 1 \) plotters use horseshoes against \( s_3 \).

6. **Calculating the total wins:**
   - The total number of wins for \( s_1 \) is \( u + 1 + \frac{2p}{3} \).
   - The total number of wins for \( s_2 \) is \( u + 1 + \frac{2p}{3} \).
   - The total number of wins for \( s_3 \) is \( u + 1 + \frac{2p}{3} \).

7. **Inequality for wins:**
   - We need the minimum number of wins obtained by \( s_4, s_5, s_6 \) to be greater than the maximum number of wins obtained by \( s_1, s_2, s_3 \):
     \[
     u + (p-3) + 1 > u + 1 + \frac{2p}{3}
     \]
     Simplifying the inequality:
     \[
     p - 2 > 1 + \frac{2p}{3}
     \]
     \[
     3(p - 2) > 3 + 2p
     \]
     \[
     3p - 6 > 3 + 2p
     \]
     \[
     p > 9
     \]

8. **Finding the minimum \( p \):**
   - Taking \( p = 10 \), we cannot reach the desired result.
   - Taking \( p = 11 \), we cannot reach the desired result.
   - Finally, \( p = 12 \) works:
     - 5 plotters use horseshoes on \( s_1 \).
     - 4 plotters use horseshoes on \( s_2 \).
     - 3 plotters use horseshoes on \( s_3 \).

Conclusion:
\[
\boxed{12}
\]