Problem 1260

1. Claim and Construction:
We claim that the maximum value of $p_1 + p_n$ is $m(n-1)$. To construct an example where this maximum is achieved, let every question be solved by the same person. In this case, the scores are:
$$p_1 = m(n-1), \quad p_2 = p_3 = \cdots = p_n = 0.$$
Thus, $p_1 + p_n = m(n-1) + 0 = m(n-1)$.

2. Optimality Proof:
Let $P_1, P_2, \ldots, P_n$ be the students, with scores $p_1, p_2, \ldots, p_n$ arranged in descending order. We need to show that:
$$p_1 + p_n \le m(n-1).$$
To do this, we will prove a stronger statement:
$$p_1 + \frac{p_2 + p_3 + \cdots + p_n}{n-1} \le m(n-1).$$
This implies that $p_1 + p_n \le m(n-1)$ because $p_n \le \frac{p_2 + p_3 + \cdots + p_n}{n-1}$.

3. Score Analysis for Each Question:
Consider a single question $q$. Let $x$ be the number of students who answered $q$ correctly. The score for each student who answered correctly is $n-x$, and the score for each student who answered incorrectly is $0$.

4. Sum of Scores for Each Question:
For each question, the total score contributed by all students is:
$$x(n-x).$$
The score $P_1$ obtained for this question is $n-x$ if $P_1$ answered correctly, and $0$ otherwise. The average score of the other $n-1$ students is:
$$\frac{(x-1)(n-x)}{n-1} \quad \text{if } P_1 \text{ answered correctly},$$
or
$$\frac{x(n-x)}{n-1} \quad \text{if } P_1 \text{ answered incorrectly}.$$

5. Bounding the Sum:
We need to show that for every question, the score $P_1$ obtained plus the average score of the other $n-1$ students is at most $n-1$. Consider the two cases:
- If $P_1$ answered correctly:
$$n-x + \frac{(x-1)(n-x)}{n-1} \le n-x + x-1 = n-1.$$
- If $P_1$ answered incorrectly:
$$0 + \frac{x(n-x)}{n-1} \le n-1.$$
Since $x(n-x) \le (n-1)(n-1)$, the inequality holds.

6. Summing Over All Questions:
Summing the inequalities over all $m$ questions, we get:
$$p_1 + \frac{p_2 + p_3 + \cdots + p_n}{n-1} \le m(n-1).$$
Therefore,
$$p_1 + p_n \le m(n-1).$$

Hence, the maximum value of $p_1 + p_n$ is $m(n-1)$.

The final answer is $\boxed{ m(n-1) }$.