Ah yes, this is such a "russian" problem, if you know what I mean.

### $\text{Partial Solution}$

Anyway, the solution is based on induction. Suppose that if we place $4n$ people at a table from $n$ countries, $4$ people from each country, we can separate them in $4$ groups such that no $2$ people from the same country are in the same group and no $2$ people sitting next to each other are from the same group.

Now, lets consider $4n+4$ people on our circle. There are $2$ cases:

### $\text{Case 1:}$

There are $2$ people from the same country that are sitting next to each other. Suppose that country is $c_i$. Take all $4$ people from country $c_i$ and ignore them. For the other $4n$ people on the circle, distribute them in $4$ groups such that the conditions are satisfied (this can be done, from the induction hypothesis).

If we prove that the $4$ people from country $c_i$ can be distributed to the $4$ groups such that the conditions in the statement still hold, then the induction is complete. Here are the $3$ possible cases of the positions of the people from country $c_i$ on the circle (here, the $X_i$s represent people from other countries and $A_i$s people from country $c_i$)

$(X_1-A_1-A_2-X_2)$, $(X_3-A_3-X_4)$, $(X_5-A_4-X_6)$

$(X_1-A_1-A_2-A_3-X_2)$, $(X_3-A_4-X_5)$

$(X_1-A_1-A_2-A_3-A_4-X_2)$

Notice that because we "ignored" the $A_i$s when distributing the $4n$ people to the $4$ groups, in each of the cases above $X_1$ and $X_2$ are in different groups; $X_3$ and $X_4$ are in different groups and $X_5$ and $X_6$ are from different groups.

It is easy to observe that, no matter the groups the $X_i$s are from, in each of the above cases the $A_i$s can be distributed in groups to still satisfy the conditions. Example: (instead of $X_i$s I just put $1$, $2$, $3$ or $4$ for the group they are in)

$(1-A_1-A_2-2)$, $(1-A_3-3)$, $(3-A_4-4)$. Take $A_3=2$, $A_4=1$, $A_1=3$, $A_2=4$ (I will not prove all the above cases, but it doesn't take much and it simply is case-work)

So in this case, the induction holds.

### $\text{Case 2:}$

There aren't $2$ people from the same country which sit next to each other. In this case we do not need induction, we will just prove that if we have $4n$ people on the circle such that there aren't $2$ people from the same country which sit next to each other we can distribute them to the $4$ groups like it was specified in the statement.

However, I haven't yet found what the algorithm is in this case. I will come back when I find it.