With Luke Robitaille and Yannick Yao. For $n=1$ the answer is $0$, so assume $n > 1$ from here. Label the couples $0, 1, \dotsc, n-1$ so that the king and queen always start with couple $0$ and the men are seated clockwise in the order $0, 1, \dotsc, n-1$. Let the women are seated clockwise in the order $0, \pi(1), \dotsc, \pi(n-1)$ where $\pi$ is a permutation of $\{1, 2, \dotsc, n-1\}$. We make the following observations: $a-1$ is equal to the number of indices $i$ with $\pi^{-1}(i) > \pi^{-1}(i+1)$ (each such index corresponds to an instance where the queen passes by woman $0$ without shaking hands). $b-1$ is equal to the number of indices $i$ with $\pi(i) > \pi(i+1)$ (each such index corresponds to an instance where the king passes by man $0$ without shaking hands). Let $s$ and $t$ be the maximum lengths of an increasing and decreasing subsequence of $\pi$, respectively. Claim 1. $t\leq a\leq n-s$. Proof. A decreasing subsequence $\pi(i_1), \dotsc, \pi(i_t)$ of length $t$ in $\pi$ corresponds to a decreasing subsequence $\pi^{-1}(\pi(i_t)), \dotsc, \pi^{-1}(\pi(i_1))$ of length $t$ in $\pi^{-1}$. Therefore $\pi^{-1}$ decreases at least $t-1$ times, so $a-1\geq t-1$. On the right hand side, an increasing subsequence of length $s$ in $\pi$ corresponds to an increasing subsequence of length $s$ in $\pi^{-1}$. Since $\pi^{-1}$ increases at least $s-1$ times, it decreases at most $n-2-(s-1)=n-1-s$ times. Therefore $a-1\leq n-1-s$. Claim 2. $t\leq b\leq n-s$. Proof. If there is a decreasing subsequence of length $t$, then $\pi$ must decrease at least $t-1$ times. Therefore $a-1\geq t-1$. On the right hand side, if there is an increasing subsequence of length $s$, then $\pi$ must increase at least $s-1$ times. Therefore $a-1\leq n-2-(s-1)=n-1-s$. Combining the two claims yields \[|a-b|\leq n-(s+t).\]Claim 3. $st\geq n-1$. Proof. This is classical. Label the $i$th element of $\pi$ with $(s_i, t_i)$ where $s_i$ is the length of the longest increasing subsequence of $\pi$ ending at the $i$th element and $t_i$ is the length of the longest decreasing subsequence of $\pi$ ending at the $i$th element. This defines an injective map $\{1, 2, \dotsc, n-1\}\to \{1, 2, \dotsc, s\}\times\{1, 2, \dotsc, t\}$. We can compute that the minimum value of $s+t$ subject to $st\geq n-1$ is $\left\lceil 2\sqrt{n-1}\right\rceil$, so \[|a-b|\leq n-\left\lceil 2\sqrt{n-1}\right\rceil.\]Finally, we provide a construction. Choose $s$ and $t$ so that $st\geq n-1$ and $s+t$ is minimal. Let \[\pi = (t, 2t, \dotsc, st, t-1, 2t-1, \dotsc, st-1, \dotsc 1, t+1, \dotsc, st-t+1),\]except with all elements greater than $n-1$ removed. (Related problem: ELMO 2016/3.)

Label the husbands in clockwise order 1, ..., n and their wives in order $\sigma(1), \cdots, \sigma(n)$. Then observe $a=\#_{1\le i\le n} \sigma(i+1)<\sigma(i)$ and $b=\#_{1\le i\le n} \sigma^{-1}(i+1)<\sigma^{-1}(i)$ The answer is $n-\lceil 2\sqrt{n-1} \rceil$ if $n\ge 2$ and $0$ otherwise. WLOG $a<b$. The key claim is $a(n-b)\ge n-1$ We can rewrite this as $a$ "chains": $$ \sigma(x_1)< \sigma(x_1+1)< \cdots < \sigma(x_2-1)$$ $$\sigma(x_2) < \sigma(x_2+1) < \cdots < \sigma(x_3-1)$$ All the way to $x_a$ to $n+x_1-1$, where $x_1<x_2<\cdots<x_a$ We need to show there are at least $\frac{n-1}{a}$ indices $j$ where $\sigma^{-1}(j+1)<\sigma^{-1}(j)$ By incrementing a value to every $x_i$ (mod n) but not changing the $\sigma$ values, $b$ doesn't change. This basically changes $\sigma^{-1} \to \sigma^{-1}+c$ and induct on $c$ to see $a$ doesn't change.) we may assume $x_1=1$. Suppose $n-b=m$ and $\sigma(x_t)=1$. This means each chain above $x_t$'s chain has length at most $m$, or observe $\sigma(x_u) < \sigma(x_u+1) < \cdots < \sigma(x_{u+1}-1) < \sigma(x_t)$ forces $n-b\ge m+1$. Similarly $x_t$'s chain has at most length $m+1$, the end! Will post construction later.