9. (1995 China National Team Selection Test for IMO) 21 people participate in an exam, which consists of 15 true/false questions. It is known that for any two people, there is at least one question that both of them answered correctly. What is the minimum number of people who answered the most answered question correctly? Please provide a justification.

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9. Let the number of people who correctly answer the $i$-th question be $a_{i}$, then the number of pairs who answer the $i$-th question correctly is
$$
b_{i}=C_{u_{i}}^{2}
$$

for $(i=1,2, \cdots, 15)$. We will focus on the sum $\sum_{i=1}^{15} b_{i}$.
Let $a=\max \left\{a_{1}, a_{2}, \cdots, a_{15}\right\}$, then we have
$$
15 C_{n}^{2} \geqslant \sum_{i=1}^{15} b_{i} \geqslant C_{21}^{2}, \quad a(a-1) \geqslant \frac{420}{15}=28, \quad a \geqslant 6 .
$$

If $a=6$, that is, at most 6 people answer each question correctly. In this case, if someone answers only three questions correctly, since $a=6$, for each of the three questions he answers correctly, at most 5 other people also answer them correctly. Therefore, he can have at most 15 people with whom he shares a correct answer, which contradicts the problem's condition. This means that in this case, each person must answer at least 4 questions correctly.
Since $21 \times 5 > 6 \times 15$,

it is impossible for everyone to answer at least 5 questions correctly, so at least one person must answer exactly 4 questions correctly. We can label this person as 1. By the problem's condition, this person must share a correct answer with each of the other 20 people, so each of the 4 questions he answers correctly must also be answered correctly by 5 other people. Thus, the people who answer these 4 questions correctly form 4 sets:
$$
\begin{aligned}
S_{1} & =\{1,2,3,4,5,6\}, S_{2}=\{1,7,8,9,10,11\}, \\
S_{3} & =\{1,12,13,14,15,16\}, S_{4}=\{1,17,18,19,20,21\} .
\end{aligned}
$$

Clearly, $S_{1} \cup S_{2} \cup S_{3} \cup S_{4}=\{1,2, \cdots, 21\}$,
$$
S_{i} \cap S_{j}=\{1\}, 1 \leqslant i<j \leqslant 4 .
$$

On the other hand, if at most 11 out of the 15 questions are answered correctly by 6 people each, then we have
$$
210=C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i} \leqslant 11 \cdot C_{6}^{2}+4 \cdot C_{5}^{2}=205,
$$

which is a contradiction. Therefore, at least 12 out of the 15 questions are answered correctly by 6 people each. Excluding the 4 questions answered correctly by the person labeled 1, at least 8 questions are answered correctly by 6 people each. Considering the 6 people who answer one of these 8 questions correctly, either 3 or more of them belong to the same $S_{j}$, or two belong to the same $S_{j}$ and two belong to the same $S_{k}, j \neq k, 1 \leqslant j \leqslant 4,1 \leqslant k \leqslant 4$. In any case, when calculating $\sum_{i=1}^{15} b_{i}$, each of these 8 questions is counted at least twice. Therefore,
$$
210=C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i}-8 \times 2 \leqslant 15 C_{i}^{2}-16=209 \text {, a contradiction. }
$$

This shows that $a \neq 6$, so $a \geqslant 7$.
The following construction shows that $a=7$ is possible.
Let the test takers be numbered $1 \sim 21$, and define
$$
\begin{array}{l}
P_{i}=\{2 i-1,2 i\}, i=1,2, \cdots, 6 ; \\
P_{7}=\{13,14,15\}, P_{8}=\{16,17,18\}, \quad P_{9}=\{19,20,21\} .
\end{array}
$$

If the test results are as shown in the table below, then $a=7$.