I believe the following schedule solves the problem as specified in 2005.
Table 1 Table 2 Table 3 Byes
(10 9 v 12 2) ( 7 11 v 5 1) (4 3 v 6 8) (0 13)
(11 10 v 13 3) ( 8 12 v 6 2) (5 4 v 0 9) (1 7)
(12 11 v 7 4) ( 9 13 v 0 3) (6 5 v 1 10) (2 8)
(13 12 v 8 5) (10 7 v 1 4) (0 6 v 2 11) (3 9)
( 7 13 v 9 6) (11 8 v 2 5) (1 0 v 3 12) (4 10)
( 8 7 v 10 0) (12 9 v 3 6) (2 1 v 4 13) (5 11)
( 9 8 v 11 1) (13 10 v 4 0) (3 2 v 5 7) (6 12)
But perhaps it might be possible to do better with the opponent balance? Here the bye pairs all oppose each other twice on table 2.
Hope that helps.