Let $\pi(i)$ denote the new location of the student originally in position $i$, where $\pi$ is a permutation of $\{1,...,n\}$. We want to maximize $$S=\sum_{i=1}^n |i-\pi(i)|=\sum_{i=1}^n(i+\pi(i)-2\min\{i,\pi(i)\}),$$ which is equivalent to minimizing $T=\sum_{i=1}^n \min\{i,\pi(i)\}$. Notice that among the numbers $1,2,\dots,n$ and $\pi(1),\pi(2),\dots,\pi(n)$, every number appears exactly twice. Hence, in $S$ each of these numbers can appear at most twice as a term in $T$. This shows that $$T\ge \begin{cases} 2(1+2+\dots+k)=k(k+1) & \text{for even }n=2k \\ 2(1+2+\dots+k)+(k+1)=(k+1)^2 & \text{for odd }n=2k+1,\end{cases}$$ and equality can be attained if we choose $(\pi(1),\pi(2),\dots,\pi(n-1),\pi(n))=(n,n-1,\dots,2,1)$. Therefore, the maximum of $S$ is $$\max S=2(1+2+\dots+n)-2T=\begin{cases} 2k(2k+1)-2k(k+1)=2k^2 & \text{for even }n=2k \\ (2k+1)(2k+2)-2(k+1)^2=2k(k+1) & \text{for odd }n=2k+1,\end{cases}$$ in other words, $\max S=\left\lfloor \frac{n^2}2\right\rfloor$.