The arithmetic does not work out on that one. With 24 players there are 276 different pairs of players.
Each foursome where no two players have played together accounts for 6 pairs. That means six
foursomes accounts for at most 36 pairs, and after 6 rounds at most 216 pairs are accounted for.
You can do pretty well for 20 players. Take the design for 25 players and delete players 21, 22, 23,
34 and 25 whenever they occur. This gives 5 days of five foursomes, and one day with four fivesomes.
There is no good way to do it with foursomes alone. The best strategy there is to select one player
from each of the fivesomes (on the last day) and let them play as a foursome. These players will have
been together before, and will not have played with the four members of the fivesome they came from.
Those are the only pairs that are duplicated and missed.
You could do a similar thing for 24 players. Start with the 25 player design and delete player 25. Each
day will have four fivesomes and a foursome. Make six foursomes by randomly choosing one person
from each fivesome (hopefully the same person would not ever get selected twice; each time someone
is selected there are three players they do not play with). This guarantees that at least 180 different pairs
play together at least once. I don't know how close you could get to 216 (the best possible) by choosing