In a hand clapping competition, the winner of each match gets 2 points, the loser gets 0 points; if it's a draw, each player gets 1 point. Now the scores are: $1979$, $1980$, $198$, $1985$. After verification, how many players participated in the competition?
Since $n(n-1)$ is the product of two consecutive natural numbers, it must be even, so 1979 and 1985 are impossible. And the last digit of $n(n-1)$, being the product of two consecutive natural numbers, can only be 0, 2, or 6, so 1984 is also impossible. The only possibility left is 1980. For $n(n-1)=1980$, solving gives $n=45$.