8. In a football tournament, 8 teams participated, and each played against each other exactly once. It is known that any two teams that drew with each other ended up with a different number of points. Find the maximum possible total number of draws in this tournament. (A team earns 3 points for a win, 1 point for a draw, and 0 points for a loss.) (S. Tokarev)
Answer: 22. Solution: We will prove that no more than two teams can have exactly 6 draws. Indeed, any such team has either 6 or $6+$ 3 points (depending on whether they won or lost their decisive match). If there are three such teams, then two of them have the same number of points, meaning they did not play to a draw against each other; this is impossible because both either did not win or did not lose any matches. It is also clear that at most one team played all 7 of their matches to a draw. Thus, the sum of the number of draws for all 8 teams does not exceed $7+6+6+5+5+5+5+5=44$, and since each draw is counted twice, the total number of draws in the tournament does not exceed $44 / 2=22$. It can equal 22, as shown in the table on the right.

|  | A | B | C | D | E | F | G | H |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| A | X | 1 | 1 | 1 | 1 | 1 | 3 | 3 |
| B | 1 | X | 1 | 1 | 1 | 1 | 1 | 3 |
| C | 1 | 1 | X | 3 | 0 | 1 | 1 | 1 |
| D | 1 | 1 | 0 | X | 3 | 1 | 1 | 1 |
| E | 1 | 1 | 3 | 0 | X | 1 | 1 | 1 |
| F | 1 | 1 | 1 | 1 | 1 | X | 1 | 1 |
| G | 0 | 1 | 1 | 1 | 1 | 1 | X | 1 |
| H | 0 | 0 | 1 | 1 | 1 | 1 | 1 | X |