The whist schedule that Richard is talking about will have 63 rounds, but you are wanting only one third of this. Lets say you took the first 21 rounds from the full whist schedule, then you would almost certainly find that some of the pairs of players in opposition, all of which occur twice somewhere in the 63 rounds, just happen to occur twice in the first 21 rounds. As a consequence you will not get the play-once-with-each-other-player property.
I can tell you that a solution to your problem does exist. Frustratingly, I have a reference book that tells me as much, but doesn't say how to construct the schedule, it just gives a reference to the original scientific paper. For completeness, this is Discrete Mathematics, vol 3 (1972), 343-357. You could
get it here, but it will cost and be warned that you may well need help turning the combinatorial math into a schedule.
