Sufficient condition for $n$ There are $n$ people (distinct men and women) sitting around the table. After the break they will sit around the table again. What is the sufficient condition for $n$ such that there always exists $2$ men or $2$ women for whom the number of people between them is the same before and after the break?
 A: This is too long for a comment, but it may help to understand the problem. Also, it may shed a light into some assumptions that may or may not be useful. 
I'm taking these assumptions:
A) It is a round table, so the $n$-tuple is cyclic. 
B) The number of men/women is not known.
C) The "pair" of men/women is made of different people. In a cyclic $n$-tuple, we may understand a pair using the repetition of elements. Without this assumption, $n\geq1$ is sufficient condition, because a pair is formed by an element and itself, so there's always the same distance between the pair, $k=n-1$. Analyze the case of $n=2$, with 1 man and one woman.
D) People between a pair is all of the same gender, different to the gender of the pair. This means that the minimal distance is sufficient to define the case-by-case sceneario. 
I'm sure this assumption can be discarded later, but I think is useful to have this assumption in the early stages of solving the problem. 

Doing a simple case analysis,
If $n=2$, two cases may happen before the break:


*

*Both persons are of same gender (2 of 4 possibilities), so they always have $k=0$ distance between them no matter the ordering.

*Both persons are of different gender (2 of 4 possibilities), so there's no distance.


After the break, the orderings will be equal. So, I am inclined to believe that $n=2$ is sufficient, because the case of different genders is not defined by the problem. 

If $n=3$, two cases may happen before the break:


*

*All people are of same gender (2 of 8 possibilities), so there's always distance $k=0$ between any pair of them.

*Two people are of one gender and the other different (6 of 8 possibilities), so there's always mínimum distance $k=0$ between the closest pair.


Note that in every case, the cases after the break will be equivalent! So, $n=3$ is sufficient: the event always happens.

If $n=4$, three cases may happen before the break:


*

*All people are of same gender (2 of 16 possibilities), so there's always distance $k=0$ between any pair of them.

*Three people are of one gender, and one is of different gender (8 of 16 possibilities), so there's always distance $k=0$ between a pair of them.

*Two men and two women (6 of 16 possibilities), so we may have distance $k=0$ or $k=1$.


So, in 10 out of 16 possibilities of men/women in the table, the event always happens. But we can't be sure in 6 of those possibilities, because the reordering may change the distance between the closest pair. Then, $n=4$ is not sufficient.

If $n=5$, three cases may happen before the break:


*

*All people are of same gender (2 of 32 possibilities), so there's always distance $k=0$ between any pair of them 

*Four people are of one gender, and one is of different gender (10 of 32 possibilities), so there's always distance $k=0$ between a pair of them. 

*Three people are of a gender, and two of the other (20 of 32 possibilities), so we may have distance $k=0$ or $k=1$.


So, in 12 out of 32 possibilities of men/women in the table, the event happens. But we can't be sure in 20 of those possibilities, because the reordering may change the distance between the closest pair.

It can be seen from the last two scenarios that everytime that the number of men is equal to the number of women ($n$ even), or a gender is equal to the other plus one ($m=w+1$ or $m=w-1$, $n$ odd), the miminum distance could be $k=0$ or $k=1$, so there's no way to impose conditions on $n$ so the event always happens. 
With this analysis, I'm inclined to believe that (under the assumptions given) the sufficient condition on $n$ to ensure the event is always true, is that $n=3$. We may also consider $n=2$, depending on the consideration of that case as valid.
Maybe the real problem is dropping assumption D). I'm trying to generalize this on paper. I'll post an answer with any findings later.
