Partition 100 people, 4 from each country into 4 groups with conditions

Here's a two-step procedure partially borrowed from this question to find a partition. (Actually, maybe it is easier to read the other way around: my explanation here feels better than the one I posted there, so people confused about my answer to that question should read this answer instead.)

Number the people around the circle $1, 2, \dots, 100$.

Step 1: let's say that each country has two men and two women representing it. Obviously it is out of our hands where the men and women will sit.

Step 2: we give everyone a color that is either Red or Blue. We will make sure that from each country, the two men get different colors, and the two women get different colors. We will also make sure that any two people numbered $2k-1$ and $2k$ for some $k$ get different colors.

Graph-theoretically, this step just means two-coloring the union of even cycles, which can always be done. Each person has two constraints: they must be different from one of the people next to them, and they must be different from the person of their country of the same gender. So the graph of constraints is $2$-regular - a union of cycles. The cycles are all even because the constraints alternate between two different types, so each cycle has an even number of edges.

Step 3: we give everyone a shade that is either Light or Dark. We will make sure that from each country, the two Reds get different shades, and the two Blues get different shades. We will also make sure that any two people numbered $2k$ and $2k+1$ for some $k$ (or numbered $100$ and $1$) get different colors.

Graph-theoretically, this problem is identical to Step 2.

Now, the partition into four groups of $25$ is as follows: the Light Red, Dark Red, Light Blue, and Dark Blue groups. By the constraints we enforced in Step 2 and Step 3, each country has one of each color, and any two people sitting next to each other disagree either in color or in shade.