Don,

I found this on the web page http://www.maa.org/editorial/mathgames/mathgames_08_14_07.html:

"28 golfers can play in foursomes for 9 days. If quadruples are picked so that every pair occurs once, S(2, 4, 28), a Steiner 2-design is the solution, and there are 4466 such solutions. From these, there are 7 different resolvable designs, two of them coming from the Ree unital. One of these solutions gives a schedule for our golfers.

Day1 ABCD EFGH IJKL MNab cdef ghij klmn
Day2 AEgk BFMc Ndhl GIem HJai CKbn DLfj
Day3 AFjn BEae bfim HKcl GLMh CINk DJdg
Day4 AIci BJNn EKMj FLdm begl CGaf DHhk
Day5 AGbd BHgm ELNi achn FKfk CJej DIMl
Day6 AKeh BLbk FIag EJfl Ncjm CHMd DGin
Day7 AHNf BGjl FJbh Meik EIdn CLcg DKam
Day8 ALal BKdi GJck Mfgn HIbj CEhm DFNe
Day9 AJMm BIfh CFil DEbc GKNg HLen adjk "

You could use it twice, changing which golfer corresponds to which letter the second time in order to get different foursomes. And do that again but use only the first six days. This is about as good as one can do. Everyone plays together 4-6 times.

Have fun!
Victoria