The answer is $\left\lceil \frac{2n+2}{3} \right\rceil$.

For $i = 0, 1, \dots, n$, let $y_i = x_i - x_{i+1}$ where $x_0 = x_{n+1} = 0$. Note that $y_0 + y_1 + \dots + y_n = 0$. The move in the problem is equivalent to the following.
Choose non-negative integers $i$ and $j$ with $0 \le i < j \le n$, and replace $y_i$ and $y_j$ by $y_i - 1$ and $y_j + 1$, respectively.
We call this move the $(i, j)$-move. Our goal is to make each $y_i$ a multiple of $3$. Now we prove that for every sequence of integers $y_0, y_1, \dots, y_n$ with $y_0 + y_1 + \dots + y_n = 0$, we can make each of $y_i$'s a multiple of $3$ by performing at most $\left\lceil \frac{2n+2}{3} \right\rceil$ moves.

We use induction on $n$.
One can easily check that it is possible when $n = 1, 2$.
Suppose $n > 2$. If $3|y_i$ for some $i$, then we can ignore $y_i$, and so we need at most $\left\lceil \frac{2n}{3} \right\rceil$ moves to make all $y_i$'s multiples of $3$. Hence, we may assume that $y_i \ne 0$ for each $i$. We consider the following four cases.

Case 1. There exist $0 \le i < j \le n$ such that $y_i \equiv 1 \pmod{3}$ and $y_j \equiv 2 \pmod{3}$.
Then, we perform the $(i, j)$-move then $y_i$ and $y_j$ become multiples of $3$. So, in this case, we need at most $1 + \left\lceil \frac{2n-2}{3} \right\rceil \le \left\lceil \frac{2n+2}{3} \right\rceil$ moves.

Case 2. There exist $0 \le i < j < k \le n$ such that $y_i \equiv y_j \equiv y_k \equiv 1 \pmod{3}$.
Then, we perform the $(i, k)$-move and the $(j, k)$-move. Then, $y_i, y_j$ and $y_k$ become multiples of $3$. So, we need at most $2 + \left\lceil \frac{2n-4}{3} \right\rceil = \left\lceil \frac{2n+2}{3} \right\rceil$ moves.

Case 3. There exist $0 \le i < j < k \le n$ such that $y_i \equiv y_j \equiv y_k \equiv 2 \pmod{3}$.
Then, we perform the $(i, j)$-move and the $(i, k)$-move. Then, $y_i, y_j$ and $y_k$ become multiples of $3$. So, we need at most $2 + \left\lceil \frac{2n-4}{3} \right\rceil = \left\lceil \frac{2n+2}{3} \right\rceil$ moves.

Case 4. Neither case 1, case 2 nor case 3 occurs.
Since $n \ge 3$, the only possible case is that $n = 3$ and $(y_0, y_1, y_2, y_3) \equiv (2, 2, 1, 1) \pmod{3}$. In this case, we perform the (0, 1)-move, the (0, 3)-move and the (2, 3)-move. Then, each $y_i$ becomes a multiple of $3$.

Therefore, by the induction, we can make each $y_i$ a multiple of $3$ by performing at most $\left\lceil \frac{2n+2}{3} \right\rceil$ moves.

Finally, for the following cases, we need at least $\left\lceil \frac{2n+2}{3} \right\rceil$ moves, which implies that the answer is $\left\lceil \frac{2n+2}{3} \right\rceil$.

$$(y_0, y_1, \dots, y_n) = \begin{cases} (2, 2, \dots, 2, 2, 2), & \text{if } n \equiv 2 \pmod{3} \\ (2, 2, \dots, 2, 2, 1), (2, 2, \dots, 2, 1, 2) & \text{if } n \equiv 1 \pmod{3} \\ (2, 2, \dots, 2, 1, 1), & \text{if } n \equiv 0 \pmod{3} \end{cases}$$