Assuming the opposite - that the partnerships are not fixed then an alternative schedule is:
round table 1 table 2 table 3 table 4 bye
1 (I O : F G) (C K : D H) (E P : R J) (B L : Q M) (A N)
2 (A P : G H) (D L : E I) (F Q : J K) (C M : R N) (B O)
3 (B Q : H I) (E M : F A) (G R : K L) (D N : J O) (C P)
4 (C R : I A) (F N : G B) (H J : L M) (E O : K P) (D Q)
5 (D J : A B) (G O : H C) (I K : M N) (F P : L Q) (E R)
6 (E K : B C) (H P : I D) (A L : N O) (G Q : M R) (F J)
7 (F L : C D) (I Q : A E) (B M : O P) (H R : N J) (G K)
8 (G M : D E) (A R : B F) (C N : P Q) (I J : O K) (H L)
9 (H N : E F) (B J : C G) (D O : Q R) (A K : P L) (I M)
Here every player, gets one bye, partners 8 different players, and opposes 16 out of the 17 other players. The player who is never opposed is always one of the 8 partners, so every possible pair of players sit together at a table at least once.
