We will show that $n = 16$.

As one can see in the figure, it is possible to arrange the numbers $1$ to $9$ at the vertices so that the sum of any three consecutive numbers is at most $16$.

Let us show that for $n < 16$ this cannot be achieved. Assume, to the contrary, that it is possible to do so. If we add the sums over all possible triplets of consecutive vertices, we have used every number three times since it occurs in three such triplets. Hence, the number we obtain is equal to $3 \cdot (1 + 2 + \dots + 9) = 135$. On the other hand, every sum of numbers at three consecutive vertices is at most $n$ and there are $9$ such sums. Thus, the total sum is at most $9n$. We conclude that $9n \ge 135$, so $n \ge 15$.

Finally, let us show that $n$ cannot equal $15$. If this were the case, all of the triplets would have to add up to $15$. Let us denote the numbers at four consecutive vertices by $a_1, a_2, a_3$ and $a_4$. Then $a_1 + a_2 + a_3 = 15$ and $a_2 + a_3 + a_4 = 15$, so $a_1 = a_4$. This is not possible since every number has to occur exactly once. This proves that $n$ has to be at least $16$.