Consider the division into arcs $A, B,C$ with sums $a\leq b \leq c$ such that $c-a$ is minimal (since there are only finitely many numbers on the circle such a division must exist). We claim that $c-a \leq 1.$ Assume otherwise, and let $x$ be the number of $C$ next to the arc $A$ and consider the division $A\cup \{x\}, B, C\setminus \{x\}.$ Then \begin{align*} |(c-x)-b| &= \max((c-x)-b, b-(c-x))\leq \max(c-b , x) \leq \max(c-a, 1) \leq c-a,\\ |b-(a+x)| &= \max(b-(a+x), a+x-b) \leq \max(b-a , x) \leq c-a,\\ |(c-x)-(a+x)| &= \max(c-a-2x, (a+x)-(c-x)) \leq \max(c-a , 2x+(a-c)) \leq \max(c-a , 2-1) \leq c-a, \end{align*}where we use that $0<x\leq1$ and note that furthermore, equality cannot arise in all equations of any line. However, this shows that the new division has a smaller maximal difference, a contradiction.