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**Grandad** What also became clear, is that each selection of 3 questions - call them questions $\displaystyle i, j$ and $\displaystyle k$ - involves the elimination of 3 pairs - $\displaystyle ij, jk, ik$ - from the $\displaystyle \tfrac{1}{2}n(n-1)$ pairs that can be chosen from the $\displaystyle n$ questions. This is because no other participant can be given any other pair of questions from $\displaystyle i, j$ and $\displaystyle k$ once this group of three has been allocated to someone.

Following up on that insight of Grandad, each triple that is selected eliminates 3 pairs of questions. Since there are $\displaystyle \tfrac{1}{2}n(n-1)$ pairs, the maximum possible number of participants is $\displaystyle \tfrac{1}{6}n(n-1)$. Grandad's computations show that this maximum is attained when n=3 and when n=7. I believe that the maximum can be attained whenever n is of the form $\displaystyle n=2^k-1$. (In fact, I'm sure of that.) I don't have a neat way of describing the construction, but I have written it out in detail for n=15, using these 35 triples:

Code:

1 2 3 2 4 6 3 4 7 4 8 12 6 8 14
1 4 5 2 5 7 3 5 6 4 9 13 6 9 15
1 6 7 2 8 10 3 8 11 4 10 14 6 10 12
1 8 9 2 9 11 3 9 10 4 11 15 6 11 13
1 10 11 2 12 14 3 12 15 5 8 13 7 8 15
1 12 13 2 13 15 3 13 14 5 9 14 7 9 12
1 14 15 5 10 15 7 10 13
5 11 12 7 11 14

I suspect that the maximum is not going to be attained except at numbers of the form $\displaystyle n=2^k-1$, and I would guess that there will not be a simple formula for the answer at a general value of n.