Three. (50 points) 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 $p_{1}+p_{n}$.
For $k=1,2, \cdots, m$, let the number of people who did not answer the $k$-th question correctly be $x_{k}$. Then the number of people who answered the $k$-th question correctly is $n-x_{k}$. According to the scoring rule, these $n-x_{k}$ people each score $x_{k}$ points on the $k$-th question.
Let the total score of the $n$ students be $S$. Then
\[
\begin{array}{l}
\sum_{i=1}^{n} p_{i}=S=\sum_{k=1}^{m} x_{k}\left(n-x_{k}\right) \\
=n \sum_{k=1}^{m} x_{k}-\sum_{k=1}^{m} x_{k}^{2} .
\end{array}
\]

Since each person can score at most $x_{k}$ points on the $k$-th question, we have
\[
p_{1} \leqslant \sum_{k=1}^{m} x_{k} \text {. }
\]

By $p_{2} \geqslant p_{3} \geqslant \cdots \geqslant p_{n}$, we know
\[
p_{n} \leqslant \frac{p_{2}+p_{3}+\cdots+p_{n}}{n-1}=\frac{S-p_{1}}{n-1} \text {. }
\]

Then $p_{1}+p_{n} \leqslant p_{1}+\frac{S-p_{1}}{n-1}$
\[
\begin{array}{l}
=\frac{n-2}{n-1} p_{1}+\frac{S}{n-1} \\
\leqslant \frac{n-2}{n-1} \sum_{k=1}^{m} x_{k}+\frac{1}{n-1}\left(n \sum_{k=1}^{m} x_{k}-\sum_{k=1}^{m} x_{k}^{2}\right) \\
=2 \sum_{k=1}^{m} x_{k}-\frac{1}{n-1} \sum_{k=1}^{m} x_{k}^{2} .
\end{array}
\]

By the Cauchy-Schwarz inequality, we have
\[
\begin{array}{l}
\sum_{k=1}^{m} x_{k}^{2} \geqslant \frac{1}{m}\left(\sum_{k=1}^{m} x_{k}\right)^{2} . \\
\text { Hence } p_{1}+p_{n} \leqslant 2 \sum_{k=1}^{m} x_{k}-\frac{1}{m(n-1)}\left(\sum_{k=1}^{m} x_{k}\right)^{2} \\
=-\frac{1}{m(n-1)}\left[\sum_{k=1}^{m} x_{k}-m(n-1)\right]^{2}+m(n-1) \\
\leqslant m(n-1) .
\end{array}
\]

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

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