Example 2 There are $r$ people participating in a chess tournament, every 2 people play one game, the winner of each game gets 2 points, the loser gets 0 points, and in the case of a draw, each gets 1 point. After the tournament, there is exactly one person who has the fewest wins, and only he has the highest score. Find the minimum value of $r$. (16th All-Russian Mathematical Olympiad problem)
Analysis and Solution: First, consider the condition: "The number of rounds where exactly one person wins is the least, and only he scores the most," let this person be $A$. To calculate $A$'s score, we can introduce parameters: suppose $A$ wins $n$ rounds and draws $m$ rounds, then $A$'s score is $2n + m$.

Next, consider the goal: "Is $r \geqslant$ ?" An obviously valid inequality is: $r \geqslant (m + n) + 1$, so we can first estimate the ranges of $m$ and $n$.

To utilize the condition "A scores the most," we need to calculate the scores of others. This requires knowing the number of wins and losses of others, which can be inferred from the condition: "A wins the fewest rounds." For any person other than $A$, they must win at least $n + 1$ rounds, thus scoring at least $2n + 2$. Therefore, $2n + 22$, which implies $m \geqslant 3$.

This estimate, though not optimal, helps us find someone with a higher score. Since $m \geqslant 3$ indicates that at least 3 people draw with $A$. Let one of these people be $B$. Then, $B$ scores at least $(2n + 2) + 1 = 2n + 3$. Thus, $2n + 33$, which implies $m \geqslant 4$.

Next, we prove that $n > 0$. In fact, if $n = 0$, meaning $A$ does not win any rounds, then $A$'s score $S(A) \leqslant r - 1$. However, the total score of $r$ people is $2 \mathrm{C}_{r}^{2} = r(r - 1)$, and each person can score on average $r - 1$. Since $A$ scores the most, $S(A) = r - 1$. This means everyone scores $r - 1$, which is a contradiction.

In summary, $r \geqslant (m + n) + 1 \geqslant 6$.
Finally, $r = 6$ is possible, as shown in the following table:
\begin{tabular}{|l|l|l|l|l|l|l|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Total Score \\
\hline$A$ & & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & & 2 & 0 & 0 & 2 & 5 \\
\hline$C$ & 1 & 0 & & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline$F$ & 0 & 0 & 0 & 2 & 2 & & 4 \\
\hline
\end{tabular}

Therefore, the minimum value of $r$ is 6.