2. 7821 people participate in an exam, with the test paper containing 15 true/false questions. It is known that any two people have at least one question that they both answered correctly. What is the minimum number of people who answered the question with the most correct answers? Please explain your reasoning.
[Solution] Let $a_i$ be the number of people who answered the $i$-th question correctly. Then, the number of pairs who answered the $i$-th question correctly is
$$b_i = C_{a_i}^2$$
for $i=1,2, \cdots, 15$.
We will focus on the sum $\sum_{i=1}^{15} b_i$.
Let $a = \max \{a_1, a_2, \cdots, a_{15}\}$, then we have
$$\begin{array}{l}
15 C_a^2 \geq \sum_{i=1}^{15} b_i \geq C_{21}^2 \\
a(a-1) \geq \frac{420}{15} = 28 \\
a \geq 6
\end{array}$$

If $a=6$, i.e., at most 6 people answered each question correctly. If there is someone who answered only three questions correctly, then since $a=6$, for each of the three questions he answered correctly, at most 5 other people also answered 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 have answered at least 4 questions correctly.

Since $21 \times 5 > 6 \times 15$,
it is impossible for everyone to have answered at least 5 questions correctly. Therefore, at least one person must have answered exactly 4 questions correctly, and 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 answered correctly must have been answered correctly by 5 other people. Thus, the people who answered these 4 questions correctly form 4 sets:
$$\begin{array}{l}
S_1 = \{1,2,3,4,5,6\}, \\
S_2 = \{1,7,8,9,10,11\}, \\
S_3 = \{1,12,13,14,15,16\}, \\
S_4 = \{1,17,18,19,20,21\}.
\end{array}$$

Clearly, $S_1 \cup S_2 \cup S_3 \cup S_4 = \{1,2, \cdots, 21\}$,
$$S_i \cap S_j = \{1\}, 1 \leq i < j \leq 4$$

On the other hand, if at most 11 out of the 15 questions were answered correctly by 6 people each, then we would have
$$210 = C_{21}^2 \leq \sum_{i=1}^{15} b_i \leq 11 \cdot C_6^2 + 4 \cdot C_5^2 = 205,$$
which is a contradiction. Therefore, at least 12 out of the 15 questions were answered correctly by 6 people each. Excluding the 4 questions answered correctly by the person labeled 1, at least 8 questions were answered correctly by 6 people each. Considering the 6 people who answered 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 $S_k$, where $j \neq k$, $1 \leq j \leq 4$, $1 \leq k \leq 4$. In any case, when calculating $\sum_{i=1}^{15} b_i$, each of these 8 questions will be counted at least twice. Therefore,
$$210 = C_{21}^2 \leq \sum_{i=1}^{15} b_i - 8 \times 2 \leq 15 C_6^2 - 16 = 209,$$
which is a contradiction.
This shows that $a \neq 6$, so $a \geq 7$.
The following construction shows that $a=7$ is possible.
Let the test takers be numbered $1 \sim 21$, and define
$$\begin{array}{l}
P_i = \{2i-1, 2i\}, i=1,2, \cdots, 6; \\
P_7 = \{13,14,15\}, \\
P_8 = \{16,17,18\}, \\
P_9 = \{19,20,21\}.
\end{array}$$

The test results are shown in the table below, where $a=7$.
\begin{tabular}{|c|c|}
\hline Question Number & Set of People Who Answered the Question Correctly \\
\hline 1 & $P_1 \cup P_2 \cup P_3$ \\
\hline 2 & $P_4 \cup P_5 \cup P_6$ \\
\hline 3 & $P_1 \cup P_4 \cup P_7$ \\
\hline 4 & $P_2 \cup P_5 \cup P_8$ \\
\hline 5 & $P_3 \cup P_6 \cup P_9$ \\
\hline 6 & $P_1 \cup P_5 \cup P_9$ \\
\hline 7 & $P_2 \cup P_6 \cup P_7$ \\
\hline 8 & $P_3 \cup P_4 \cup P_8$ \\
\hline 9 & $P_1 \cup P_6 \cup P_8$ \\
\hline 10 & $P_2 \cup P_4 \cup P_9$ \\
\hline 11 & $P_3 \cup P_5 \cup P_7$ \\
\hline 12 & $P_7 \cup P_8$ \\
\hline 13 & $P_8 \cup P_9$ \\
\hline 14 & $P_7 \cup P_9$ \\
\hline 15 & $\varnothing$ (empty) \\
\hline
\end{tabular}