$n$ students take a test with $m$ questions, where $m,n\ge 2$ are integers. The score given to every question is as such: for a certain question, if $x$ students fails to answer it correctly, then those who answer it correctly scores $x$ points, while those who answer it wrongly scores $0$. The score of a student is the sum of his scores for the $m$ questions. Arrange the scores in descending order $p_1\ge p_2\ge \ldots \ge p_n$. Find the maximum value of $p_1+p_n$. 
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 \( \boxed{ m(n-1) } \).