It is not possible to have a solution where all pairs of players meet at least twice within a foursome, the best that can be done is to have each player play three times with 2 of their opponents and only play once with 1 other opponent (the other 4 opponents are seen exactly twice). For example:
(3 5 4 1) (2 8 6 7)
(5 7 3 8) (4 2 1 6)
(3 4 8 6) (7 1 5 2)
(8 3 1 2) (6 7 4 5)
(5 8 2 4) (1 6 7 3)
