It works for all $n \geq 1$, and can be constructed by near 1-factorisations. (I'm going to assume the "$6n+1$ professors" is a typo for $6n+4$, when all the seats are filled. If not, then just kick three people out of the conference, and the result is still true.)
Label the professors $x$ and $y_{ij}$ for $1 \leq i \leq 2n+1$ and $1 \leq j \leq 3$.
Let $F=\{F_1,F_2,\ldots,F_{2n+1}\}$ be a near 1-factorisation of $K_{2n+1}$ (which exist for all $n \geq 1$; see my answer here for a proof). We will interpret each $F_i=\{a^{(i)}_t b^{(i)}_t\}_{t=1}^{2n}$ to be a set of $2n$ edges between vertices in $\{1,2,\ldots,2n+1\}$.
At meeting $i \in \{1,2,\ldots,2n+1\}$:
At the 4-seater table we seat $\{x\} \cup \{y_{ij}\}_{j=1}^3$ where $i$ is the element in $\{1,2,\ldots,2n+1\}$ that is not in an edge in $F_i$.
At the $t$-th 6-seater table we seat $\{y_{a^{(i)}_t j}\}_{j=1}^3$ in the three "even" seats $\{y_{b^{(i)}_t j}\}_{j=1}^3$ in the three "odd" seats.
If two professors sit next to, or adjacent from, each other more than once, then $F$ is not a near 1-factorisation.
Here's a drawing for the $n=1$ case:
n=1 example.">