We claim that the answer is $\tfrac{2}{3}n$ when $3 \mid n$ and $0$ otherwise. To show this, let $a_1, \dots, a_n$ be the numbers in clockwise order, with $a_n$ the maximum of the numbers. Note that all of the numbers are nonnegative. Claim: $a_n \leq \max\{a_1, a_2\}$. Proof: Suppose not. Let $i$ be the smallest index such that $a_i > \max\{a_1, a_2\}$. (Clearly $i \geq 3$.) But then $\max\{a_1, a_2\} < a_i = |a_{i-1} - a_{i-2}| \leq \max\{a_{i-1}, a_{i-2}\}$ (where the last step follows from the fact that $a_{i-1}$ and $a_{i-2}$ are nonnegative). This implies either $a_{i-1}$ or $a_{i-2}$ is greater than $\max\{a_1, a_2\}$, contradicting our assumption that $i$ was minimal. Since we assumed $a_n$ was the maximum, it follows that either $a_1 = a_n$ or $a_2 = a_n$. In either case, we find that the numbers repeat in a pattern of period $3$: $$a_n, a_n, 0, a_n, a_n, 0, \dots$$We can have $a_n \neq 0$ if and only if $3 \mid n$ (otherwise, the pattern won't line up once you wrap around the circle). If $3 \mid n$, we can maximize the sum by taking $a_n = 1$, which gives a sum of $\tfrac{2}{3}n$. If $3 \nmid n$, then our only choice is to take $a_n = 0$, giving a sum of $0$, as claimed.