We can in fact prove that there is always three successive integers in this arrangement, whose sum is at least $33$. Though it is not clear if $33$ is the optimal lower bound.
Consider the circular arrangement of numbers starting from $1$ as follows. $1 , a_1, a_2, \ldots, a_{19}$ where $a_1,a_2,\ldots a_{19} \in \{2,3,4,\ldots,20\}$.
Note that $1 + a_1 + a_2 + \cdots a_{19} = 210$. Note that at least one of $a_1,a_4,a_7,a_{10},a_{13},a_{16},a_{19}$ must be $\leq 14$.
Say $a_1 \leq 14$, then we get that $1 + a_1 + a_2 + \cdots a_{19} \leq 1 + 14 + a_2 + \cdots a_{19}$. Let $s$ be the maximum possible sum of three consecutive elements. Then we have that $(a_2 + a_3 + a_4) + (a_5 + a_6 + a_7) + \cdots +(a_{17} + a_{18} + a_{19}) \leq 6s$
Hence, we get that $210 = 1 + a_1 + a_2 + \cdots a_{19} \leq 1 + 14 + a_2 + \cdots a_{19} \leq 6s + 15$ i.e. $6s \geq 195 \implies s \geq 32.5.$ Hence, $s \geq 33.$
The same argument works if $a_1 > 14$ and one of $a_4,a_7,a_{10},a_{13},a_{16},a_{19} \leq 14$. For instance, if $a_7 \leq 14$, then rearrange the sum as $1 + (a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + a_7 + (a_8 + a_9 + a_{10}) + (a_{11} + a_{12} + a_{13}) + (a_{14} + a_{15} + a_{16}) + (a_{17} + a_{18} + a_{19}).$