Unfortunately, there is no fair solution to this problem. Please read
this thread for more info.
About the best you can do is something like this:
( 9 8 5 6) (12 10 11 4) (3 2 1 7)
(10 5 6 7) ( 9 11 12 1) (4 3 2 8)
(11 6 7 8) (10 12 9 2) (1 4 3 5)
(12 7 8 5) (11 9 10 3) (2 1 4 6)
where 12 pairs of teams never meet and 18 pairs of teams play twice.