We will in fact prove that there is always three successive integers in this arrangement, whose sum is at least $18$.
Consider the circular arrangement of numbers starting from $1$ as follows. $1 , a_1, a_2, \ldots, a_9$ where $a_1,a_2,\ldots a_9 \in \{2,3,4,\ldots,10\}$.
Note that $1 + a_1 + a_2 + \cdots a_9 = 55$.
If $(a_1 + a_2 + a_3) \leq 17$ and $(a_4 + a_5 + a_6) \leq 17$ and $(a_7 + a_8 + a_9) \leq 17$, then $1 + a_1 + a_2 + \cdots a_9 \leq 52$. Which is clearly not the case, since they add up to $55$.
Hence, at least one of $(a_1 + a_2 + a_3)$ or $(a_4 + a_5 + a_6)$ or $(a_7 + a_8 + a_9)$ is greater than $17$.
$18$ is in fact the optimal lower bound on the sum which can be seen by considering the following arrangement: $\{1,10,6,2,8,7,3,5,9,4\}$ arranged circularly. The sum of any three taken in a circular fashion is at most $18$.