Ok, more thoughts... the odd case is trivial in the algorithmic sense. The pattern I've printed previously for N=7 should work for any odd N. The first half of the field of the even case is also trivial in the algorithmic sense. Player 1 will always have N-1 .. 2, Player 2 will be N-2 to 3, N-1, 1, Player 3 will be N-3 .. (N-1) .. 1, N-2, N-3 .. and all other players from 4 down to N/2 will be the same.
in reply to Algorithm problem: Order matches by difference between players
The second half of the even case gets complicated, because in manual tests of the case of 8 I saw situations where patterns had to be adjusted because of prior use of a pairing.