Problem 1797

For any $k = 1, 2, \dots, m$, assuming there are $x_k$ students failing to answer the $k$th question correctly, then there are $n - x_k$ ones who answer it correctly and each gets $x_k$ marks from it accordingly. Suppose the sum of the $n$ students' total marks is $S$. Then we have
$$\sum_{i=1}^{n} p_i = S = \sum_{k=1}^{m} x_k (n - x_k) = n \sum_{k=1}^{m} x_k - \sum_{k=1}^{m} x_k^2.$$
As each student gets at most $x_k$ marks from the $k$th question, we have
$p_1 \le \sum_{k=1}^{m} x_k.$
Since $p_2 \ge \cdots \ge p_n$, then $p_n \le \frac{p_2 + p_3 + \cdots + p_n}{n-1} = \frac{S - p_1}{n-1}$.
Therefore,
$$\begin{align*} p_1 + p_n &\le p_1 + \frac{S-p_1}{n-1} = \frac{n-2}{n-1}p_1 + \frac{S}{n-1} \\ &\le \frac{n-2}{n-1} \cdot \sum_{k=1}^{m} x_k + \frac{1}{n-1} \cdot \left( n \sum_{k=1}^{m} x_k - \sum_{k=1}^{m} x_k^2 \right) \\ &= 2 \sum_{k=1}^{m} x_k - \frac{1}{n-1} \cdot \sum_{k=1}^{m} x_k^2. \end{align*}$$
By the Cauchy Inequality, we have
$$\sum_{k=1}^{m} x_k^2 \ge \frac{1}{m} \left( \sum_{k=1}^{m} x_k \right)^2 .$$
Then
$$\begin{aligned} p_1 + p_n &\le 2 \sum_{k=1}^{m} x_k - \frac{1}{m(n-1)} \cdot \left( \sum_{k=1}^{m} x_k \right)^2 \\ &= -\frac{1}{m(n-1)} \cdot \left( \sum_{k=1}^{m} x_k - m(n-1) \right)^2 + m(n-1) \\ &\le m(n-1). \end{aligned}$$
On the other hand, if there is a student who answers all the questions correctly, while the other $n-1$ students fail to answer any questions, then we have
$$p_1 + p_n = p_1 = \sum_{k=1}^{m} (n-1) = m(n-1).$$
Therefore, the maximum possible value of $p_1 + p_n$ is $m(n-1)$. $\boxed{}$