4. There are several international chess teams participating in an international chess competition. It is stipulated that winning a game earns 2 points, drawing a game earns 1 point, and losing a game earns 0 points. It is also stipulated that players from different teams play one game against each other, while players from the same team do not play against each other. The statistical results of the competition show that there are a total of 9 players from all teams, a total of 23 games were played, and each player's score is different. The player with the highest score won all their games, the sum of the scores of the second and fourth place players is 3 points more than the score of the first place player, and the score of the third place player is equal to the total score of the last four players. Try to answer:
(1) How many international chess teams participated in this competition, and how many players did each team have?
(2) What are the scores of the top five players? Explain your reasoning.
Solution: (1) According to the solution of Question 3, there are three chess teams in this competition, each with 5, 3, and 1 players, respectively.
(2) The 9 players in total played 23 games, with a total score of 46 points.

Since the team with 5 players, each played 4 games, a player who won all games would score 8 points; the team with 3 players, each played 6 games, a player who won all games would score 12 points; the team with only 1 player, that player played 8 games, and if that player won all games, the score would be 16 points. Therefore, the scores for players who won all their games are 8, 12, and 16 points.

Let the scores of the top five players be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\left(a_{1}>a_{2}>a_{3}>a_{4}>a_{5}\right)$. Thus,
$$
\begin{array}{l}
a_{2}+a_{4}=a_{1}+3, \\
a_{1}+a_{2}+2 a_{3}+a_{4}+a_{5}=46 .
\end{array}
$$

Therefore, $2 a_{1}+2 a_{3}+a_{5}=43$.
Since each game can only result in scores of 0, 1, or 2, $a_{5}$ must be an odd number. Also, the minimum scores for the last four players are 0, 1, 2, and 3, so $a_{5} \geqslant 5, a_{3} \geqslant 7$.
$1^{\circ}$ When $a_{1}=16$,
$$
2 a_{3}+a_{5}=11 \Rightarrow 3 a_{5}a_{5}$, so we get $a_{5}=5, a_{3}=7$. At the same time, $a_{2}+$ $a_{4}=15$, since $a_{5}<a_{4}<a_{3}$, so $a_{4}=6$, thus $a_{2}=9$.

In conclusion, the scores of the top five players are 12, $9, 7, 6, 5$.