suppose question $i$ was answered correctly by $t_i$ students, we must have: $\sum\binom{t_i}{2}$ is at least $210=\binom{21}{2}$, so let $T=max(t_1,t_2...,t_{15})$, then $15\binom{T}{2}$ is at least 210. So $\binom{T}{2}$ is at least 14, so T is at least $6$. I dont find a construction right now....

I don't think $T=6$ is possible. Assume each question was solved by at most 6 people. Then each person must have solved at least 4 questions (since each correct question is shared with at most 5 other people). Furthermore, if anyone solved exactly 4 questions there must be 20 other solvers of those questions and they must all be distinct. Since there are 15 questions, there were at most 90 correct answers total by all participants on all questions, so at least 15 people must have solved exactly 4 questions. Let $A_1, A_2, \dots A_{15}$ be 15 people answering only 4 questions. Each pair of them must have solved exactly one question together. If any question was solved by 5 of the $A_i$, then those 5 people would have had to solve 15 other questions between them, which would all have to be distinct. This is impossible, so each question was solved by at most 4 of the $A_i$. We now have that $A_1$ solved exactly 4 questions, each of which was solved by at most 3 other $A_i$. Thus $A_1$ solved questions with at most 12 other $A_i$, contradicting that every pair of people solved a common problem.

I also realised that during the night: Like we have equality every where, we must have every pair appears together in exactly $1$ problem, and also, every one appears in exactly $5$ problems, Now to everyone asign a vector with $1$ in the $j$th place if he solved problem $j$, and $0$ otherwise. Call this vectors $v_1,v_2...,v_{21}$ and $v_iv_j=1$ and $v_iv_i=5$. Now there is a linear combination of them that sum $0$, taking products wit all vectors we find a contradiction!

To finish things off, here's a construction with $T=7$ and 14 questions overall Label the contestants by points $(i,j)$, $1 \leq i \leq 7$, $1 \leq j \leq 3$ Let one question be solved by all contestants of the form $(i, ai+b)$ for each $1 \leq a \leq 3$, $1 \leq b \leq 3$. The 10th question is solved by all contestants where $i \in \{2, 5\}$ The 11th question is solved by all contestants where $i \in \{3, 6\}$ The 12th question is solved by all contestants where $i \in \{1, 4\}$ The 13th question is solved by all contestants where $i \in \{1, 7\}$ The 14th question is solved by all contestants where $i \in \{4,7\}$

$T=7$,13 questions is enough. {1,2,3,4} {1,5,6,7} {1,8,9,10} {1,11,12,13} {2,5,8,11}*2 {2,6,9,12}*2 {2,7,10,13}*2 {3,5,9,13}*2 {3,6,10,11}*2 {3,7,8,12}*2 {4,7,9,11}*2 {4,6,8,13}*2 {4,5,10,12}

kevinatcausa wrote: Since there are 15 questions, there were at most 90 correct answers total by all participants on all questions, so at least 15 people must have solved exactly 4 questions. I don't understand this

Pascual2005 wrote: I also realised that during the night: Like we have equality every where, we must have every pair appears together in exactly $1$ problem, and also, every one appears in exactly $5$ problems, Now to everyone asign a vector with $1$ in the $j$th place if he solved problem $j$, and $0$ otherwise. Call this vectors $v_1,v_2...,v_{21}$ and $v_iv_j=1$ and $v_iv_i=5$. Now there is a linear combination of them that sum $0$, taking products wit all vectors we find a contradiction! It is impossible for everyone to be in exactly 5 problems.