The third question: An exam consists of $m$ questions, with $n$ students participating, where $m, n \geqslant 2$ are given integers. The scoring rule for each question is: if exactly $x$ students fail to answer the question correctly, then each student who answers the question correctly gets $x$ points, and those who fail to answer correctly get zero points. Each student's total score is the sum of their scores on the $m$ questions. The total scores of all students are arranged in descending order as $p_{1} \geqslant p_{2} \geqslant \cdots \geqslant p_{n}$. Find the maximum possible value of $p_{1}+p_{\mathrm{n}}$.
Solution 1 First, use mathematical induction to prove the inequality $p_{1}+p_{n} \leqslant m(n-1)$ for a positive integer $m$.
(1) When $m=1$, if the student who gets $p_{n}$ points answers this question correctly, then the other students should also answer this question correctly.
At this time, $p_{1}=p_{2}=\cdots=p_{n}=0$.
If the student who gets $p_{n}$ points does not answer this question correctly, it is obvious that $p_{1}+p_{n}=p_{1}+0 \leqslant n-1$.
(2) If it holds for $m$ questions, we have
$p_{1}+p_{n} \leqslant m(n-1)$.
For $m+1$ questions, we have
(i) The student with the lowest score answers all $m+1$ questions correctly, then the other students also answer all $m+1$ questions correctly. Thus,
$p_{1}=p_{2}=\cdots=p_{n}=0$.
(ii) The student with the lowest score fails to answer at least one question, arbitrarily select a question that is not answered correctly and denote it as $a$, and the remaining $m$ questions as set $A$. Let the score sequence of $n$ people in $A$ be $p_{1}^{\prime} \geqslant p_{2}^{\prime} \geqslant \cdots \geqslant p_{n}^{\prime}$, and the score sequence in $m+1$ questions be $p_{1} \geqslant p_{2} \geqslant \cdots \geqslant p_{n}$. Since the student who gets $p_{n}$ points does not answer $a$ correctly, we still have $p_{n}=p_{n}^{\prime}$.
By the induction hypothesis, $p_{1}^{\prime}+p_{n}^{\prime} \leqslant m(n-1)$.
Notice that, if the student with the highest score in $A$ does not answer $a$ correctly, then the highest scorer in $m+1$ questions may change, but we still have $p_{1} \leqslant p_{1}^{\prime}+(n-1)$. Thus,
$$
\begin{array}{l}
p_{1}+p_{n} \leqslant p_{1}^{\prime}+p_{n}+n-1 \\
=p_{1}^{\prime}+p_{n}^{\prime}+n-1 \\
\leqslant(m+1)(n-1) .
\end{array}
$$

If the student who gets $p_{1}^{\prime}$ points answers $a$ correctly, the score of this question is less than or equal to $n-1$, so $p_{1} \leqslant p_{1}^{\prime}+n-1$, that is,
$$
p_{1}+p_{n} \leqslant(m+1)(n-1) \text {. }
$$

By mathematical induction, we know that for all positive integers,
$$
p_{1}+p_{n} \leqslant m(n-1) \text {. }
$$

On the other hand, if one student answers all questions correctly and the other $n-1$ students answer all questions incorrectly, then
$$
p_{1}+p_{n}=p_{1}=\sum_{k=1}^{m}(n-1)=m(n-1) \text {. }
$$

In summary, the maximum value of $p_{1}+p_{n}$ is $m(n-1)$.
Solution 2 Let the $n$ students be $A_{1}, A_{2}, \cdots, A_{n}$, and the total score of $A_{i}$ be $p_{i}(1 \leqslant i \leqslant n)$. Suppose $A_{1}$ gets $x_{i}(0 \leqslant i \leqslant n-1)$ questions with $i$ points. Thus,
$$
x_{0}+x_{1}+\cdots+x_{n-1}=m,
$$

and $p_{1}=x_{1}+2 x_{2}+\cdots+(n-1) x_{n-1}$.
For the question where $A_{1}$ gets $i(1 \leqslant i \leqslant n-1)$ points, then exactly $i$ students did not answer it correctly, so exactly $n-i$ students answered it correctly. Thus, the total score of students $A_{2}, A_{3} \cdots, A_{n}$ on this question is $(n-i-1) i$. For the question where $A_{1}$ gets 0 points, suppose exactly $x$ students did not answer it correctly. Then the score of students $A_{2}$, $A_{3}, \cdots, A_{n}$ on this question is
$$
\begin{array}{l}
x(n-x) \leqslant \frac{n^{2}}{4} . \\
\text { Hence } p_{1}+p_{n} \leqslant p_{1}+\frac{p_{2}+p_{3}+\cdots+p_{n}}{n-1} \\
\leqslant x_{1}+2 x_{2}+\cdots+(n-1) x_{n-1}+ \\
\frac{\frac{n^{2}}{4} x_{0}+(n-2) x_{1}+2(n-3) x_{2}+\cdots+(n-2) x_{n-1}}{n-1} \\
=\sum_{i=1}^{n-1}\left[i+\frac{i(n-i-1)}{n-1}\right] x_{i}+\frac{n^{2}}{4(n-1)} x_{0} \\
\leqslant \sum_{i=1}^{n-1}(i+n-i-1) x_{i}+(n-1) x_{0} \\
=(n-1) m .
\end{array}
$$

On the other hand, if one student answers all questions correctly and the other $n-1$ students answer all questions incorrectly, then
$$
p_{1}+p_{n}=\sum_{k=1}^{m}(n-1)=m(n-1) .
$$

In summary, the maximum value of $p_{1}+p_{n}$ is $m(n-1)$.