$n$ people sitting on a circular table without repeating neighbour-sets
$\def\FF{\mathbb{F}}\def\PP{\mathbb{P}}$I can achieve the bound $\binom{n-1}{2}$ if $n=q+1$ for a prime power $q$.
Outline of method: Suppose that $X$ is a set of size $n$ and $G$ is a group acting on $X$ in a sharply three transitive manner. This means that, is $(x_1, x_2, x_3)$ and $(y_1, y_2, y_3)$ are two ordered triples of distinct elements of $G$, there is precisely one $g \in G$ with $g(x_i)=y_i$ for $i=1$, $2$, $3$. (So, in particular, we have $|G| = n(n-1)(n-2)$.) Suppose furthermore that there is an element $\gamma$ in $G$ which acts by an $n$-cycle on $X$, and $\gamma$ is conjugate to $\gamma^{-1}$.
Then I claim $n$ is achievable. Let $C \subset G$ be the conjugacy class of $\gamma$. We note that the only permutations in $S_n$ which commute with an $n$-cycle are powers of that $n$-cycle, so $Z(\gamma) = \langle \gamma \rangle$ and the size of the conjugacy class $C$ is $|G|/|Z(\gamma)| = n(n-1)(n-2)/n=(n-1)(n-2)$.
I claim that, for any distinct elements $y_1$, $y_2$, $y_3$ in $X$, there is a unique $\delta \in C$ with $\delta(y_1) = y_2$ and $\delta(y_2) = y_3$. We first see that there is at least one $\delta$: Fix $x_1 \in X$, define $x_2 = \gamma(x_1)$ and $x_3 = \gamma(x_2)$, and let $h \in G$ be such that $h(x_i) = y_i$. Then $\delta = h \gamma h^{-1}$ does the job. But $|C| = (n-1) (n-2)$ and each $\delta \in C$ gives rise to $n$ triples $(y_1, y_2, y_3)$ with $\delta(y_1) = y_2$, $\delta(y_2) = y_3$. So, if each triple occurs at most once, then each triple occurs exactly once.
We have shown that the elements of $C$ are $(n-1)(n-2)$ oriented $n$-cycles, with each $y_1 \to y_2 \to y_3$ occurring in exactly one. But also, we assumed $\gamma$ conjugate to $\gamma^{-1}$, so for every cycle in $C$ its reverse is also in $C$. Identifying these, we get $\binom{n-1}{2}$ unoriented cycles, where for each $y_2$, and each pair of neighbors $y_1$, $y_3$, there is exactly one cycle where they occur.
Details Take $X = \PP^1(\FF_q)$ and $G = PGL_2(\FF_q)$ with the obvious action. It is well known that $G$ is sharply triply transitive. Choose an identification of $(\FF_q)^2$ with the field $\FF_{q^2}$, choose a generator $\theta$ for the cyclic group $\FF_{q^2}^{\times}$ and let $\gamma \in GL_2(\FF_q)$ be the matrix of multiplication by $\theta$. Then the image of $\gamma$ in $PGL_2(\FF_q)$ has order $q+1$. Moreover, in $PGL_2(\FF_q)$, every element is conjugate to its inverse. This proves the claim. $\square$.
Zassenhaus classified the sharply triple transitive permutation groups and, for all of them, $|X|=q+1$ for a prime power $q$. So there aren't any other $n$ achievable in this way.
I prove the following:
- The OP's claim follows from the perfect-1-factorization conjecture: $K_{2m}$ can be decomposed into $2m-1$ perfect matchings such that the union of any two matchings forms a Hamiltonian cycle of $K_{2m}$.
An equivalent version is that $K_{2m-1}$ can be decomposed into $2m-1$ near-perfect matchings such that the union of any two near-perfect matchings forms a Hamiltonian path. Here, a near perfect matching is a collection of edges which induce a matching that covers all but one vertex.
- If $n$ is even, then we can make $\dfrac{(n-1)\varphi(n-1)}{2}$ configurations, where $\varphi$ is the Euler totient function. [Corrected from previous wrong claims.]
Proof of 1: Given the conjecture for a particular even $n$, we take the union of every pair of matchings to be the configuration.
Proof of 2: Let $N=n-1$. We identify one person as $N$ and the others with $\{0,1,\ldots,N-1\}$ and work in arithmetic modulo $N$.
For each pair $(a,b)$ with $0 \leq a< b < N$, we define a configuration $C_{a,b}$ in which the neighbors of $x$ are $a-x$ and $b-x$. Also notice that when $N$ is odd, both $a/2$ and $b/2$ really have only one neighbor, therefore in this case, we make them both adjacent to $N$ in $C_{a,b}$ (idea of David Speyer).
We have considered the union of the two matchings $\{(x,a-x)\}$ and $\{(x,b-x)\}$. In general, this union need not be a hamiltonian cycle (which we want a configuration to be).
Claim: If $gcd(a-b,N)=1$, then $C_{a,b}$ is a valid cyclic permutation.
To prove the claim, let's consider the following edge coloring of the complete graph $K_N$: the edge $\{x,y\}$ is labeled by $x+y$. Notice that this is a proper edge coloring, i.e. no two edges get the same color. Now consider the subgraph induced by two color classes, $a,b$. Suppose that this subgraph has a cycle, say $x_1,x_2,\ldots,x_{2k}$. Then $x_1-x_3=a-b,\ldots,x_{2k-1}-x_1=a-b$. Adding these up, we get: $k(a-b)=0$. Since $k<N$, we must have $gcd(a-b,N)>1$. This shows that when $gcd(a-b,N)=1$, the union of the two color classes $a,b$ forms a Hamiltonian path between $a/2$ and $b/2$.
Counting: The number of configurations is $\sum_{gcd(d,N)=1} (N-d)=N\varphi(N)/2$.