I think 6n+4 mathematicians is more adequate?

Oh, yep, I am sorry for my blindness It must be $ 6n+4$ instead of $ 6n+1$. I will tell Orlando to edit the problem statement. Thank you!

I think that problem is quite easy. Both answer are "yes" Divide $ 6n+4$ people into $ 2n+2$ "teams": one team of one member(for convenience we call him $ A$), and $ 2n+1$ teams of three member. It is obvious that we can arrange a tournament of $ 2n+1$ rounds such that any two teams exactly meet once in the tournament. Now we use the arrangement of $ 2n+1$ rounds to arrange the $ 2n+1$ meetings: In each round there is a team meeting $ A$. Let these four people sitting at the 4-person table; other $ 2n$ teams take part in $ n$ matches. For any two these teams meeting each other, let the six people sitting at the 6-person table such that everyone's both neighbors are of the other team. Finally check that this arrangement satisfies the condition. We claim that for any two people, they sit next to or have opposite places exactly once.(for convenience we call this relationship "meet") For $ A$ and another one, their unique "meet" occurs when they both sit at the 4-person table. For any two people other than $ A$, if they are of the same team, their unique "meet" occurs when they both sit at the 4-person table. if they are of different teams, their unique "meet" occurs when their teams meet. In other words, it occurs when their two teams sit at one 6-person table together. So our claim is proved and our arrangement is enable. Q.E.D.