It's gonna be really tough

According to my understanding, I think that it should be 6 3+2 10 = 38 it should be 10 pairs when contestants solve 5 problems

I am going to make a claim, and say that any solution to the problem would need to have all people solving 4 or 5 problems. I think it is strictly better in all occasions to have participants solve at least 4 problems. It would be interesting if someone could either prove or disprove this claim. However it seems correct enough, so I'm working with it.

If you have N participants, then you need each couple of problems to be solved by at least 2/5 N + 1 participants. We will divide the participants into those that solve 4 problems (X) and the rest (N - X).

Assume X = N. Then each person solves 4 problems. By symmetry*, you can assume the solved problems to be equally distributed: there are 6 over 4 possible choices (=15), and we have 1/15 X participants solving each quadruple of problems. This means each individual couple is solved by all those participants solving those two problems plus any other two problems, i.e. 4 over 2 possible categories, which is 6. Naturally, 6/15 = 2/5. So right now, with an equal distribution, we have that all couples of problems are solved by exactly 40% of the participants.

Here is where it gets interesting. The correct solution is said to be 2. However, if you have N-X = 2, then these people each do not solve a problem. Let these problems that are not solved be problem A and problem B. Then the couple AB is solved only by 2/5 X people, which is less than 2/5 N people. All couples solved by at least one of the two smarter people is solved by more than 40% of people, but there is one couple that isn't.

If you think this way, then the smallest number of 5-problems-people required is 4. 3 smart people cover all couples at least once more, but (2/5 X +1) / X + 3 is not 40% (obvious, 33.3% < 40%). So you need 4 smart people, which cover all couples at least 2 times between them.

The problem I have with the above reasoning (simply counting solved couples) is that it doesn't take into account the fact that you can't distribute couples as you prefer between all participants. Couples are "grouped together" based on the actual problems solved: you can't have one person solve couple AB, couple CD, couple EF, couple AE, couple EF, couple BD for example: they are six unique couples, but they cover 6 distinct problems!

The problem I have with my solution is the symmetry step, marked with an asterisk * above. Is it possible to distribute couples to the X participants so that, with only two smart participants, you can have each couple solved 2/5 X + 1 times?

The above mentioned grouping of couples leads me to believe you cannot. There is exactly one couple that is problematic in the 2 smart people solution, and if you try to add one more "copy" of that couple in the rest of the participants, then they will lose some couples and gain some more to adjust to the new solved problem, which will simply shift the heat to another couple (or more).

Overall, I am not convinced by the solution "2", but I am not thoroughly convinced 4 is needed either.