Austrian Mathematical Olympiad

All the numbers are absolute values, so they are positive or zero. Either all of them are zero, then their sum is also zero, or there is a maximal positive number. If we scale all numbers such that this maximum is $1$, the sum can only get larger, therefore, we may assume that the maximum is $1$ in this case.

We observe that the difference of two such neighboring numbers smaller than $1$ is also smaller than $1$. If we iterate around the circle, we see that all numbers have to be smaller than $1$ which is impossible if the maximum is $1$.

Therefore, in the case of maximum $1$, at least one of each pair of neighbors must equal $1$. We can now distinguish two cases for the two numbers after the maximum. Either the number immediately after the maximum is also $1$ which means that the list of differences continues as $1,1,0,1,1,0,1,1,0,\dots$ or the next numbers are $1,x,1$. However this means that $|1-x|=1$ so that $x=0$ and we get the list $1,0,1,1,0,1,1,0,1,\dots$.

In both these subcases, $n$ has to be divisible by $3$ to wrap correctly around the circle with these patterns. Then, we get a sum of $2n/3$.

In conclusion, we get maximal sum $0$ if $n$ is not divisible by $3$ and $2n/3$ if $n$ is divisible by $3$.