Choose $3n$ points on a circle, show that there are two diametrically opposite point

Well, that's simple. If we want to avoid having opposite points, then obviously every 1-arc must be positioned diametrically opposite to the middle of some 3-arc. Consider one such pair. What is the length of the great arc connecting their edges? Apparently, $3n-2$, which means it has the same parity as $n$. But it is composed of $n-1$ odd arcs (1s or 3s) and some unknown number of 2-arcs which do not matter, so it must have the opposite parity.

That's it.

Assume otherwise:

On a circle of length $6n$, we choose $3n$ points such that they split the circle into $n$ arcs of length $1$, $n$ arcs of length $2$, $n$ arcs of length $3$; and no two points are diametrically opposite.

Let's remove every arc of length $1$ from the circle, and also remove its opposite segment, which is the middle of an arc of length $3$. This operation preserves pairs of opposite points, and it simplifies the setup to this:

On a circle of length $4n$, we choose $2n$ points such that they split the circle into $2n$ arcs of length $2$; and no two points are diametrically opposite.

Now the contradiction is clear: starting from any point, traverse $n$ arcs of length $2$.