Find the maximum number of draws in a tournament.

It's not possible to have more than one team draw all their matches.
It's not possible for more than one team to have six draws and one win (two such teams must have drawn their match, since neither lost). Similarly, it's not possible for more than one team to have six draws and one loss, or for more than one team to have five draws and two wins, or five draws and two losses.
It's possible for the remaining three teams to have five draws, one win and one loss, but this can only happen if none of the matches between these three teams are drawn.

This means that at least six non-draws are required. Actually, all the above can happen simultaneously and you should be able to find a way to achieve this.

[edit to include an example, since another answer claims this is impossible]

Say A draws all their matches (7 points), B beats E but draws all other matches (9 points), C beats D and E but draws the rest (11 points), D loses to C but draws the rest (6 points), E loses to B and C but draws the rest (5 points), F beats G but loses to H (8 points), G beats H but loses to F (8 points), H beats F but loses to G (8 points).

Every pair of teams which drew their match finish on a different number of points. F and G finish on the same number, but didn't draw, and neither did G and H, or F and H.