Solutions for four have been posted.
Three is impossible, because each musician must either:
- play alone (and have all 5 listen)
- play at least twice (if they are not playing alone, they must play for musician(s) playing with them the first time).
They must also either
- be in the audience alone (and listen to all five)
- be in the audience at least twice (they must listen to the person who was in the audience with them the first time)
For everyone to listen twice and play twice, 24 'slots' are needed, but only 18 are available.
If one plays alone, then in the remaining two concerts, the 5 others must all play twice, which means they cannot listen again.
If one listens alone, then in the remaining two concerts, the 5 others must all listen twice, which means they cannot play again.
Therefore it is impossible with three concerts.
There is no solution for 4 concerts other than for them all to be 3 vs 3, because:
Nobody can listen to anybody more than twice or play for anybody more than twice, there are only 24 slots, and we've already established 24 are required.
So suppose ABCD played to EF.
A would still need to play to B and C and D, and EF cannot listen to A again, so AEF to BCD is the next concert.
B also needs to play to ACD. Now E and F have both played to C and heard C, so cannot do either again without going over 24 spaces.
So no 4 vs 2 concerts.
It does not matter in fact whether E was listening in the first concert or playing, so the 5 vs 1 concert is out as well.