If you have been looking for a 6 round schedule where every pair of players oppose at least once, then I am sorry to say that this is impossible - it is a mathematically proven fact. It is possible to come close, a 6 round schedule exists (not shown) where only one pair of players does not meet - perhaps you found this already?
For the general problem with 3 players per board, some numbers of players like 9 and 15 work perfectly while others like 12 and 18 do not. I have put the smallest schedule for 12 players below where all pairs oppose at least once. Here the 1st round is short, only requiring one board.
(1 2 3) ( - - -) (- - -) (- - -)
(2 5 10) ( 7 4 11) (6 3 8) (1 9 12)
(3 6 11) ( 8 5 12) (4 1 9) (2 7 10)
(1 4 12) ( 9 6 10) (5 2 7) (3 8 11)
(6 12 7) (11 1 10) (8 2 9) (5 3 4)
(4 10 8) (12 2 11) (9 3 7) (6 1 5)
(5 11 9) (10 3 12) (7 1 8) (4 2 6)
Clearly, players 1 to 3 have one more match than the others, and as a consequence have 3 repeated opponents each, while players 4 to 12 only have one repeated opponent each. When you use the schedules you may choose not to play these repeated matches and give the players a bye.