For any even integer $n \geqslant 4$, we consider the case
$$a_{1}=a_{2}=\cdots=a_{n-1}=b_{n}=0 \quad \text { and } \quad b_{1}=b_{2}=\cdots=b_{n-1}=a_{n}=1$$
The condition $\left|a_{k}\right|+\left|b_{k}\right|=1$ is satisfied for each $1 \leqslant k \leqslant n$. No matter how we choose each $x_{k}$, both sums $\sum_{k=1}^{n} x_{k} a_{k}$ and $\sum_{k=1}^{n} x_{k} b_{k}$ are odd integers. This implies $\left|\sum_{k=1}^{n} x_{k} a_{k}\right| \geqslant 1$ and $\left|\sum_{k=1}^{n} x_{k} b_{k}\right| \geqslant 1$, which shows (1) cannot hold.
For any odd integer $n \geqslant 3$, we may assume without loss of generality $b_{k} \geqslant 0$ for $1 \leqslant k \leqslant n$ (this can be done by flipping the pair $\left(a_{k}, b_{k}\right)$ to $\left(-a_{k},-b_{k}\right)$ and $x_{k}$ to $-x_{k}$ if necessary) and $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m} \geqslant 0>a_{m+1} \geqslant \cdots \geqslant a_{n}$. We claim that the choice $x_{k}=(-1)^{k+1}$ for $1 \leqslant k \leqslant n$ will work. Define
$$s=\sum_{k=1}^{m} x_{k} a_{k} \quad \text { and } \quad t=-\sum_{k=m+1}^{n} x_{k} a_{k} .$$
Note that
$$s=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{4}\right)+\cdots \geqslant 0$$
by the assumption $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m}$ (when $m$ is odd, there is a single term $a_{m}$ at the end, which is also positive). Next, we have
$$s=a_{1}-\left(a_{2}-a_{3}\right)-\left(a_{4}-a_{5}\right)-\cdots \leqslant a_{1} \leqslant 1$$
Similarly,
$$t=\left(-a_{n}+a_{n-1}\right)+\left(-a_{n-2}+a_{n-3}\right)+\cdots \geqslant 0$$
and
$$t=-a_{n}+\left(a_{n-1}-a_{n-2}\right)+\left(a_{n-3}-a_{n-4}\right)+\cdots \leqslant-a_{n} \leqslant 1 .$$
From the condition, we have $a_{k}+b_{k}=1$ for $1 \leqslant k \leqslant m$ and $-a_{k}+b_{k}=1$ for $m+1 \leqslant k \leqslant n$. It follows that $\sum_{k=1}^{n} x_{k} a_{k}=s-t$ and $\sum_{k=1}^{n} x_{k} b_{k}=1-s-t$. Hence it remains to prove
$$|s-t|+|1-s-t| \leqslant 1$$
under the constraint $0 \leqslant s, t \leqslant 1$. By symmetry, we may assume $s \geqslant t$. If $1-s-t \geqslant 0$, then we have
$$|s-t|+|1-s-t|=s-t+1-s-t=1-2 t \leqslant 1$$
If $1-s-t \leqslant 0$, then we have
$$|s-t|+|1-s-t|=s-t-1+s+t=2 s-1 \leqslant 1$$
Hence, the inequality is true in both cases. These show $n$ can be any odd integer greater than or equal to 3.