China Hong Kong Mathematical Olympiad

The answer is $2 \cdot 3^n - 2^n$.
Let $y_j = x_{2j-1} + x_{2j}$ for $j = 1, 2, \dots, n$. Every desired sequence $\{x_j\}$ corresponds to a sequence $\{y_j\}$ such that $y_j = -2, 0, 2$ for each $j$, and $|y_k + y_{k+1} + \dots + y_m| \le 2$ for any $1 \le k \le m \le n$. Suppose there are $r$ terms of $\{y_j\}$ which are nonzero, say $y_{i_1}, y_{i_2}, \dots, y_{i_r}$ where $i_1 < i_2 < \dots < i_r$. Since
$$|y_{i_1} + y_{i_1+1} + \dots + y_{i_2}| = |y_{i_1} + y_{i_2}| \le 2,$$
we must have $y_{i_2} = -y_{i_1}$. Similarly, $y_{i_3} = -y_{i_2} = y_{i_1}$, and inductively $y_{i_{2j-1}} = y_{i_1}$ and $y_{i_{2j}} = -y_{i_1}$ for any integer $j$. In other words, the sequence $\{y_j\}$ only depends on the indices $i_1, i_2, \dots, i_r$ and the term $y_{i_1}$.
Conversely, for any such sequence $\{y_j\}$, consider any $1 \le k \le m \le n$. Let $i_s$ be the smallest index such that $k \le i_s$, and let $i_t$ be the largest index such that $i_t \le m$. Then
$$y_k + y_{k+1} + \dots + y_m = y_{i_s} + y_{i_{s+1}} + \dots + y_{i_t} = \pm(2 - 2 + 2 - 2 + \dots) = -2, 0, 2.$$
So the condition is satisfied.
Now, for each sequence $\{y_j\}$ satisfying the condition, we can construct the sequence $\{x_j\}$ as follows. For any index $j$ of the form $i_u$, the terms $x_{2j-1}$ and $x_{2j}$ are uniquely determined. Indeed, if $y_{i_u} = \pm 2$, then $x_{2i_u-1} = x_{2i_u} = \pm 1$. For any other index $j$, there are two choices for $x_{2j-1}$ and $x_{2j}$, since exactly one of them is 1 and the other is -1. Since there are $\binom{n}{r}$ ways to choose the indices $i_1, i_2, \dots, i_r$ and two ways to choose $y_{i_1}$ if $r \ge 1$, the total number of sequences is
$$\binom{n}{0}2^n + \sum_{r=1}^{n} 2\binom{n}{r} \cdot 2^{n-r} = 2^n + 2\sum_{r=1}^{n} \binom{n}{n-r} 2^{n-r} = 2^n + 2\sum_{d=0}^{n-1} \binom{n}{d} 2^d$$
using the change of variable $d = n-r$. By the binomial theorem, this is equal to
$$2^n + 2 \sum_{d=0}^{n} \binom{n}{d} 2^d - 2^{n+1} = 2^n + 2(2+1)^n - 2^{n+1} = 2 \cdot 3^n - 2^n.$$