3. A basketball championship has been played in a double round-robin system (each pair of teams play each other twice) and without ties (if the game ends in a tie, there are overtimes until one team wins). The winner of the game gets 2 points and the loser gets 1 point. At the end of the championship, the sum of the points obtained by all the teams except the champion is 2015 points. How many games has the champion won?
Solution. Suppose the number of teams is $n$. Then, a total of $2\left(\begin{array}{l}n \\ 2\end{array}\right)=n^{2}-n$ matches are played in the championship due to the double round-robin format. In each match, 3 points are awarded, so $3 n^{2}-3 n$ is the total number of points awarded. If the champion has $P$ points, and the other $n-1$ teams have a total of 2015 points, then

$$
P=3 n^{2}-3 n-2015
$$

where additionally $P>\frac{2015}{n-1}$ to be the champion. For this to hold, it must be

$$
3 n^{2}-3 n-2015>\frac{2015}{n-1}, \quad 3 n(n-1)>\frac{2015 n}{n-1}, \quad n-1>\sqrt{\frac{2015}{3}}
$$

Since $25^{2}=62526$, or $n \geq 27$.

On the other hand, the maximum score the winner could have obtained is $4(n-1)$, if they won all their matches (2 matches with each of the other $n-1$ teams), that is,

$$
3 n^{2}-3 n-2015 \leq 4(n-1), \quad(3 n-4)(n-1) \leq 2015
$$

Now, if $n \geq 28$, then $3 n-4 \geq 80, n-1 \geq 27$, and $80 \cdot 27=2160>2015$, so it must be $n \leq 27$.

Therefore, $n=27$ is the number of teams in the championship, the number of points obtained by the champion is $3 \cdot 27^{2}-3 \cdot 27-2015=91$, and since these points were obtained in $2 \cdot 26=52$ matches, the number of matches won (in which 2 points are awarded instead of 1) is clearly $91-52=39$.