3. 21 people participate in an exam, with the test paper containing 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 questions correctly? Please explain your reasoning.
For the $i$-th problem, $a_{i}$ people answered correctly, thus there are exactly
$$
b_{i}=C_{s_{i}}^{z}
$$

pairs of people who answered the $i$-th problem correctly $(i=1,2, \cdots, 15)$.
Below, we 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
$$
15 C_{a}^{2} \geqslant \sum_{i=1}^{15} b_{i} \geqslant C_{21}^{2}, \quad a(a-1) \geqslant \frac{420}{15}=28,
$$
$a \geqslant 6$.
We will show that $a=6$ is impossible, hence, the problem with the most correct answers must have at least 7 people answering correctly.

Assume $a=6$, meaning no more than 6 people answered any question correctly. We will derive a contradiction. If someone answered only three questions correctly, then for each of these questions, they could only share the correct answer with at most 5 other people, meaning they could share correct answers with at most 15 other people, which contradicts the assumption. Therefore, each person must answer at least four questions correctly.
Since $21 \times 5 > 6 \times 15$,

it is impossible for everyone to answer five or more questions correctly, so there must be at least one person who answers exactly four questions correctly, and these four questions form four sets that have no common members except for this person, for example:
$$
\begin{array}{l}
S_{1}=\{1,2,3,4,5,6\}, \\
S_{2}=\{1,7,8, \cdots, 11\}, \\
S_{3}=\{1,12,13, \cdots, 16\}, \\
S_{4}=\{1,17,18, \cdots, 21\},
\end{array}
$$

On the other hand, among all 15 questions, at least 12 questions must have 6 people answering correctly, otherwise it would lead to a contradiction:
$$
C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i} \leqslant 11 C_{6}^{2}+4 C_{5}^{2}=205 .
$$

Excluding the aforementioned four questions, there are still eight questions with 6 people answering correctly. Consider the 6 people who answered one of these eight questions correctly. Among them, either more than three people belong to the same $S_{j}$, or two people belong to some $S_{j}$, and two others belong to $S_{k}(j, k \in\{1,2,3,4\}, j \neq k)$. In any case, when calculating $\sum_{i=1}^{15} b_{i}$, each of these eight questions will result in at least two duplicate counts. This leads to
$$
C_{21}^{2} \leqslant \sum_{i=1}^{15} b_{i}-8 \times 2 \leqslant 15 C_{6}^{2}-16=209 .
$$

The resulting contradiction shows that the case $a=6$ is impossible.
Based on the above discussion, we confirm that $a \geqslant 7$.
Next, we construct an example to show that the case $a=7$ is possible. For this, we first number the participants from 1 to 21 and define the following sets of participants:
$$
\begin{array}{l}
P_{i}=\{2 i-1,2 i\}, i=1, \cdots, 63 \\
P_{1}=\{13,14,15\}, \\
P_{8}=\{16,17,18\}, \\
P_{0}=\{19,20,21\},
\end{array}
$$

Using these notations, we construct the following table to indicate the participants who answered each question correctly. It is easy to verify that in the scenario shown in the table, every pair of the 21 participants has at least one question they both answered correctly, and no more than 7 people answered the same question correctly.