2. In an international chess tournament, winning a game earns 2 points, a draw earns 1 point, and losing a game earns 0 points. Now, there are 10 players participating in a round-robin tournament (each pair of players plays one game). After all the games are played, it is found that all players have different scores. When the players are ranked by their scores from highest to lowest, the first and second place players have not lost any games. The sum of the scores of the first and second place players is 20 points more than the score of the second place player. It is also known that the fourth place player's score is equal to the total score of the last four players. What are the scores of the top six players? Explain your reasoning.
(1999, Shaoxing Tianma Cup Mathematics Competition)
Solution: Let the scores of the top six players be $a_{1}, a_{2}$, $a_{3}, a_{4}, a_{5}, a_{6}$.

Since 10 players participate in the competition, each playing 9 games, and the top two players have no losses, and the scores are only 2, 1, 0, we have
$$
a_{1} \leqslant 17, a_{2} \leqslant 16 \text {. }
$$

This competition involves $\frac{9 \times 10}{2}=45$ games, with a total score of 90 points. Therefore, we have
$$
\begin{array}{l}
a_{1}+a_{2}=a_{3}+20, \\
2 a_{3}+2 a_{4}+a_{5}+a_{6}=90-20=70 . \\
a_{1} \leqslant 17, a_{2} \leqslant 16 \\
\Rightarrow a_{3}+20 \leqslant 33 \\
\Rightarrow a_{3} \leqslant 13 . \\
a_{3}>a_{4}>a_{5}>a_{6} \\
\Rightarrow 6 a_{3} \geqslant 70 \Rightarrow a_{3} \geqslant 12 .
\end{array}
$$

Thus, $12 \leqslant a_{3} \leqslant 13$, meaning $a_{3}$ can only be 12 or 13.
$$
\begin{array}{l}
a_{3}=12 \Rightarrow 2 a_{4}+a_{5}+a_{6}=46 \\
\Rightarrow 4 a_{4}>46 \Rightarrow a_{4}>11 \Rightarrow a_{4}=12 .
\end{array}
$$

This is incorrect.
When $a_{3}=13$, on one hand, we know
$$
a_{1}=17, a_{2}=16 \text {; }
$$

On the other hand,
$$
\begin{array}{l}
2 a_{4}+a_{5}+a_{6}=44 \Rightarrow 4 a_{4}>44 \\
\Rightarrow a_{4}>11 \Rightarrow a_{4}=12 \Rightarrow a_{5}+a_{6}=20 .
\end{array}
$$

Since $a_{5}>a_{6}$, we have $a_{5}=11, a_{6}=9$.
In summary, the scores of the top six players are 17,
$$
16,13,12,11,9 \text {. }
$$