Answer: If $n = m^2 - d, 0 \le d < m$, then the maximum value is $(n - 2m)$; if $n = m^2 - m - d, 0 \le d < m - 1$, then the maximum value is $(n - 2m + 1)$.

- Estimation: Without loss of generality, assume $a > b$. Let the starting points of the host gentleman and hostess lady be the reference points of the two tables, numbered clockwise $0 \sim n-1$ in order, and let the wife of gentleman $i$ be numbered $x_i$. Considering the situations in both methods when the host gentleman and hostess lady pass by $0$, we obtain:
$$a - 1 = |\{i \in [n-1] \mid x_i > x_{i+1}\}|$$
$$b - 1 = |\{i \in [n-1] \mid x_j = i, x_l = i+1, l < j\}|$$
(where $[n-1] := \{1, 2, \dots, n-1\}$)

Let $x_1, \dots, x_{r_1}; x_{r_1+1}, \dots, x_{r_2}; \dots; x_{r_{k-1}+1}, \dots, x_{r_k}$ be $k$ decreasing sequences with $x_{r_t} < x_{r_{t+1}}$, then we have $a-1=n-k$.

Consider treating $[n-1]$ as points arranged in order; if $x_j = i, x_l = i+1, l > j$, then connect $j$ and $l$ with a line segment. This diagram will form some chains (each point connects to at most one line on the left and at most one on the right), and the leftmost end of a chain is precisely an $l$ satisfying $x_j = i, x_l = i+1, l < j$. However, the number of chains is greater than or equal to $r_{t+1} - r_t$ (for all $t=1, 2, \dots, k$), because $r_t + 1, \dots, r_{t+1}$ are not connected to each other, and thus we obtain

$$b-1 \ge \left\lfloor \frac{n-1}{k} \right\rfloor.$$
Furthermore, we obtain:
$$a-b \le n-k - \left\lfloor \frac{n-1}{k} \right\rfloor$$
If $n = m^2 - d, 0 \le d < m$, then $a - b \le n - 2m$.
If $n = m^2 - m - d, 0 \le d < m - 1$, then $a - b \le n - 2m + 1$.

- Construction:

* If $n = m^2 - d, 0 \le d < m$, then let $\{x_{r_{t-1}+1}, x_{r_{t-1}+2}, \dots, x_{r_t}\}$ be the decreasing sequence formed by the numbers in $[n-1]$ congruent to $t$ modulo $m$. Then in this case $k=m$, and the number of chains is also $m$ (the chain heads being $1, m+1, \dots, m^2-2m+1, m^2-m+1$ respectively), so in this case $|a-b| = n-2m$.

* If $n = m^2 - m - d, 0 \le d < m-1$, then let $\{x_{r_{t-1}+1}, x_{r_{t-1}+2}, \dots, x_{r_t}\}$ be the decreasing sequence formed by the numbers in $[n-1]$ congruent to $t$ modulo $m$. Then in this case $k=m$, and the number of chains is $m-1$ (the chain heads being $1, m+1, \dots, m^2-2m+1$ respectively), so in this case $|a-b| = n - 2m + 1$.