Consider $y \in B$, arbitrarily chosen. Denote by $A_y$ the number of pairs $(x, z) \in A \times C$ for which $x < y < z$, and by $B_y$ the number of pairs $(x, z) \in A \times C$ for which $x > y > z$.

Let $A = \{a_1 < a_2 < \dots < a_n\}$ and $C = \{c_1 < c_2 < \dots < c_n\}$.
On the real axis, $y$ separates the numbers $a_1, a_2, \dots, a_n$ into $k$ numbers smaller than $y$ and $n-k$ numbers greater than $y$, where $k$ is some number from $\{0, 1, 2, \dots, n\}$. Likewise, $c_1, c_2, \dots, c_n$ are separated by $y$ into $p$ numbers smaller than $y$ and $n-p$ numbers greater than $y$, where $p \in \{0, 1, 2, \dots, n\}$. It follows that $A_y = k(n-p)$ and $B_y = p(n-k)$, so $A_y - B_y = n(k-p)$.

Summing over all $y \in B$, we get $a - b = \sum_{y \in B} (A_y - B_y) = n \sum_{y \in B} (k-p)$. Since $k$ and $p$ depend on $y$, but for each $y$ the difference $k-p$ is an integer, and the sum is multiplied by $n$, it follows that $a - b$ is divisible by $n$.