I hope the contest will be interesting for both divisions. There will be 12 problems.

I see that there are some people that agree with me(judging by upvotes), so more might have faced this issue. May one of the contest writers clarify the statement or give some explanation of a sample?

Hint

Consider building a solution inductively (we claim that the answer is always YES).

If $$$n = 3$$$, then there is always a cyclic order (clockwise or anticlockwise) along which the $$$a_i$$$'s are non-decreasing. Hence, using that, we can easily construct a solution for $$$n = 3$$$. Note that in this solution, no set is empty.

We can look at the assignment as assigning arrows between $$$a_i$$$ and $$$a_{i+1}$$$ (indices wrap around), where the clockwise arrows are for assigning the conversation to player 1 and anticlockwise ones for assigning the conversation to player 2. The "balance" of the configuration can be defined as the sum of $$$to - from$$$ over all arrows $$$(from, to)$$$, and it is sufficient to find a balanced assignment of arrows inductively.

Now consider how we get from $$$i-1$$$ to $$$i$$$. You need to fit $$$a_i$$$ between $$$a_1$$$ and $$$a_{i-1}$$$. To maintain the balance between the two players, you need to assign directions to arrows between $$$a_{i-1}$$$ and $$$a_i$$$, and $$$a_i$$$ and $$$a_1$$$ depending on the direction of the arrow that is currently between $$$a_{i-1}$$$ and $$$a_1$$$.

As it turns out, this can be done in a manner similar to the $$$n = 3$$$ case, and you'll be done inductively (in $$$O(n)$$$).