4. In a chess tournament, every two players play exactly one game against each other. The winner of each game gets 2 points, the loser gets 0 points, and in the case of a draw, both players get 1 point. Now, there are 4 scorers who have tallied the total points of the tournament, but due to some being careless, their data are different, namely 1979, 1980, 1984, 1985. After verification, one of them is correct. The total number of players in this tournament is $\qquad$.
[Answer] 45
[Solution] It is easy to know that regardless of whether the match is a win or loss situation, or a draw, the total score of two people in one game is always 2 points. Therefore, the final total score should be an even number. From the four answers, it is clear that 1984 or 1980 might be the total score. This means the total number of matches is $1984 \div 2=992$ matches or $1980 \div 2=990$ matches. Suppose there are $n$ players in the competition, and each pair of players plays exactly one match, then the total number of matches should be $C_{n}^{2}=\frac{n(n-1)}{2}$. So we need to see if $\frac{n(n-1)}{2}=992$ has a solution or $\frac{n(n-1)}{2}=990$ has a solution. It is easy to know that $\frac{45 \times 44}{2}=990$, so the answer is 45