8.4. In how many ways can all natural numbers from 1 to $2 n$ be arranged in a circle so that each number is a divisor of the sum of its two neighboring numbers? (Ways that differ only by rotation or symmetry are considered the same)
Solution. Note that if the numbers in the circle do not alternate in parity, then some two even numbers are adjacent. An even number is a divisor of the sum of the even number adjacent to it and the second adjacent number, which means the latter is also even. Continuing this reasoning around the circle, we get that all numbers in the circle are even, which is incorrect. Consider the number $2n-1$. By the proven fact, both of its neighbors are even, and their sum is divisible by $2n-1$, so the sum is at least $4n-2$. The maximum sum of the two remaining numbers is $2n + (2n-2) = 4n-2$. Therefore, the numbers adjacent to $2n-1$ are $2n$ and $2n-2$. Without loss of generality, let them be in descending order clockwise. Now and further, the situation will be as follows: around the circle, the numbers from $2n$ to $k$ are already placed in descending order clockwise, where $k>1$. One of the neighbors of the number $k$ is $k+1$, and the other neighbor is no more than $k-1$, so the sum of the adjacent numbers is greater than $k$ and no more than $2k$, which means it is equal to $2k$. It turns out that the next number after $k$ is $k-1$. We again get the situation described above. Therefore, the numbers are arranged in a circle from $2n$ to 1 in descending order clockwise or counterclockwise (these are the same options) and no other way.

Answer. 1 option.