Al,
The smallest number of rounds for which it is possible to have a schedule with the equal number of times property is 23, at which point all pairs of players occur together within foursomes exactly 3 times. For example:
(24 1 8 11) (2 9 13 23) (3 6 7 12) (4 10 15 19) (5 17 18 20) (14 16 21 22)
(24 2 9 12) (3 10 14 1) (4 7 8 13) (5 11 16 20) (6 18 19 21) (15 17 22 23)
(24 3 10 13) (4 11 15 2) (5 8 9 14) (6 12 17 21) (7 19 20 22) (16 18 23 1)
(24 4 11 14) (5 12 16 3) (6 9 10 15) (7 13 18 22) (8 20 21 23) (17 19 1 2)
(24 5 12 15) (6 13 17 4) (7 10 11 16) (8 14 19 23) (9 21 22 1) (18 20 2 3)
(24 6 13 16) (7 14 18 5) (8 11 12 17) (9 15 20 1) (10 22 23 2) (19 21 3 4)
(24 7 14 17) (8 15 19 6) (9 12 13 18) (10 16 21 2) (11 23 1 3) (20 22 4 5)
(24 8 15 18) (9 16 20 7) (10 13 14 19) (11 17 22 3) (12 1 2 4) (21 23 5 6)
(24 9 16 19) (10 17 21 8) (11 14 15 20) (12 18 23 4) (13 2 3 5) (22 1 6 7)
(24 10 17 20) (11 18 22 9) (12 15 16 21) (13 19 1 5) (14 3 4 6) (23 2 7 8)
(24 11 18 21) (12 19 23 10) (13 16 17 22) (14 20 2 6) (15 4 5 7) (1 3 8 9)
(24 12 19 22) (13 20 1 11) (14 17 18 23) (15 21 3 7) (16 5 6 8) (2 4 9 10)
(24 13 20 23) (14 21 2 12) (15 18 19 1) (16 22 4 8) (17 6 7 9) (3 5 10 11)
(24 14 21 1) (15 22 3 13) (16 19 20 2) (17 23 5 9) (18 7 8 10) (4 6 11 12)
(24 15 22 2) (16 23 4 14) (17 20 21 3) (18 1 6 10) (19 8 9 11) (5 7 12 13)
(24 16 23 3) (17 1 5 15) (18 21 22 4) (19 2 7 11) (20 9 10 12) (6 8 13 14)
(24 17 1 4) (18 2 6 16) (19 22 23 5) (20 3 8 12) (21 10 11 13) (7 9 14 15)
(24 18 2 5) (19 3 7 17) (20 23 1 6) (21 4 9 13) (22 11 12 14) (8 10 15 16)
(24 19 3 6) (20 4 8 18) (21 1 2 7) (22 5 10 14) (23 12 13 15) (9 11 16 17)
(24 20 4 7) (21 5 9 19) (22 2 3 8) (23 6 11 15) (1 13 14 16) (10 12 17 18)
(24 21 5 8) (22 6 10 20) (23 3 4 9) (1 7 12 16) (2 14 15 17) (11 13 18 19)
(24 22 6 9) (23 7 11 21) (1 4 5 10) (2 8 13 17) (3 15 16 18) (12 14 19 20)
(24 23 7 10) (1 8 12 22) (2 5 6 11) (3 9 14 18) (4 16 17 19) (13 15 20 21)
Is that too many rounds for you?
Ian.