Some thoughts.
Using $-\arccos u = \arccos(-u) - \pi$ on $[-1, 1]$, we have
$$\sum_{i = 1}^n \Big(\arccos(x_i) - \arccos(y_i)\Big)
= -n\pi + \sum_{i=1}^n \Big(\arccos(x_i) + \arccos( -y_i)\Big).$$
Thus, equivalently we turn to find the maximum of
$$-n\pi + \sum_{i=1}^{2n} \arccos z_i$$
subject to $z_i\in [-1, 1], \forall i$
and $\sum_{i=1}^{2n} z_i = 0$.
Let $f(u) := \arccos u$. Note that $f(u)$ is convex on $[-1, 0]$, and concave on $[0, 1]$.
To proceed, we need the following auxiliary result. For more details, see Bin Zhao's half concave half convex theorem. Also, see my answer here.
Theorem 1. Let $a, b, c$ be fixed reals ($a < c < b$), and let $x_1, x_2, \cdots, x_n \in [a, b]$ ($n\ge 2$) with $x_1 + x_2 + \cdots + x_n = S$ (constant).
Let $f$ be a continuous function on $[a, b]$, convex on $[a, c]$ and concave on $[c, b]$. Then $f(x_1) + f(x_2) + \cdots + f(x_n)$ is maximal for $x_1 = \cdots = x_{k-1} = a$ and $x_{k+1} = \cdots = x_n$
for some $k \in \{1, 2, \cdots, n\}$.
By Theorem 1 ($a=-1, c = 0, b = 1$), $-n\pi + \sum_{i=1}^{2n} \arccos z_i$ is maximal for $z_1 = \cdots = z_{k-1} = -1$, and $z_k \in [-1, 1]$, and $z_{k+1} = \cdots = z_{2n} = \frac{k - 1 - z_k}{2n - k} \in [-1, 1]$
for some $k \in \{1, 2, \cdots, n+1\}$.
Moreover, if $k = n + 1$,
then from $\frac{k - 1 - z_k}{2n - k}\in [-1, 1]$, we have $z_k = 1$, $z_{k+1} = \cdots = z_{2n} = 1$,
and $-n\pi + \sum_{i=1}^{2n} \arccos z_i = 0$. Thus, we can assume that $k \in \{1, 2, \cdots, n\}$.
Thus, equivalently we turn to solve
$$\max_{z \in [-1, 1],\, 1 \le k \le n} \left[-n\pi + (k-1)\pi + \arccos z
+ (2n-k)\arccos \frac{k - 1 - z}{2n - k}\right]. \tag{1}$$
Let
$$F(z) := -n\pi + (k-1)\pi + \arccos z
+ (2n-k)\arccos \frac{k - 1 - z}{2n - k}.$$
We have
$$F'(z) = 0, \quad z \in (-1, 1) \implies z \in \left\{ - \frac{k-1}{2n - 1 - k},\quad \frac{k-1}{2n+1-k}\right\}. \tag{2}$$
Furthermore, we have
$$F(-1) - F\left(- \frac{k-1}{2n - 1 - k}\right)
$$
$$= (2n-k)\arccos\frac{k}{2n-k}
- (2n-1-k)\arccos\frac{k-1}{2n-1-k} \ge 0, \tag{3}$$
and
$$F\left(\frac{k-1}{2n+1-k}\right) - F(1) $$
$$ = (2n+1-k)\arccos\frac{k-1}{2n+1-k}
- (2n-k)\arccos\frac{k-2}{2n-k}\ge 0. \tag{4}$$
Using (2)-(4), (1) is equivalent to
$$\max_{1 \le k \le n, \, z = -1, \frac{k-1}{2n+1-k}} \left[-n\pi + (k-1)\pi + \arccos z
+ (2n-k)\arccos \frac{k - 1 - z}{2n - k}\right],$$
which is further equivalent to
$$\max_{0 \le k \le n} \left[-n\pi + k\pi
+ (2n-k)\arccos \frac{k}{2n - k}\right]. \tag{5}$$
Remark. From (5), $-n\pi + \sum_{i=1}^{2n} \arccos z_i$ is maximal for $z_1 = \cdots = z_k = -1,\, z_{k+1} = \cdots = z_{2n} = \frac{k}{2n-k}$ for some $k \in \{0, 1, \cdots, n\}$. Back to the original problem, the maximizer is the form of $(x_1, \cdots, x_n) = (-1, \cdots, -1, c, \cdots, c)$ and $(y_1, \cdots, y_n) = (-c, \cdots, -c)$ for some $c\in (0,1)$.
When $n=8$, the maximizer is $(x_1, \cdots, x_8) = (-1, -1, \cdots, -1, 3/5, 3/5), \quad (y_1, \cdots, y_8) = (-3/5, \cdots, -3/5)$.
