5. A circle has a regular 2018-gon inscribed in it. The numbers 1, 2, ..., 2018 are arranged at the vertices of the 2018-gon, with one number per vertex, such that the sum of any two adjacent numbers (from this arrangement) is equal to the sum of the two numbers diametrically opposite to them. Determine the number of different arrangements of the numbers.

(Arrangements obtained by rotation around the center of the circle are considered the same).
Solution. We consider an arrangement of the numbers that satisfies the condition of the problem. Let $A, B$ be two adjacent numbers in this arrangement and their diametrically opposite numbers be $a, b$ respectively. Then the condition $A+B=a+b$ or $A-a=b-B$ is satisfied. From the arbitrariness of the adjacent numbers $A, B$, we get that the difference of the numbers at diametrically opposite positions is a constant, i.e., $A-a=C$. Therefore, the numbers from 1 to 2018 need to be grouped into 1009 pairs such that the difference between the numbers in each pair is $C$. Such pairs can be formed if and only if $C$ is a divisor of 1009.

Let $C=k$. Then (correspondingly) the numbers must be paired as follows:

\[
\begin{aligned}
& \{1,2, \ldots, k\} \rightarrow\{k+1, k+2, \ldots, 2 k\} \\
& \{2 k+1,2 k+2, \ldots, 3 k\} \rightarrow\{3 k+1,3 k+2, \ldots, 4 k\}, \ldots \\
& \{2019-2 k, \ldots, 2017-k, 2018-k\} \rightarrow\{2019-k, \ldots, 2017,2018\}
\end{aligned}
\]

($k$ elements from one set are paired with $k$ elements from another set, with each element of the set $\{1,2, \ldots, 2018\}$ appearing exactly once).

If the number of (arrow) pairings in (1) is $m$, then $2018=2 \mathrm{~km}$ or $1009=k m$. Since 1009 is a prime number, the only possible values for $k$ are 1 and 1009.

1) $C=1$. In this case, the pairing of the numbers will be $(1,2),(3,4)$, ..., $(2017,2018)$, and the numbers in each pair will be diametrically opposite. Let's fix the first pair (1,2) on one diagonal. Note that it is necessary to choose the order of the numbers between 1 and 2 in one direction (for example, in the direction of the clock hands), and the other half will be determined by the condition of the problem. Since rotations are not counted, the adjacent number to 1 (in the direction of the clock hands) can be chosen in 1008 ways, its adjacent in 1007 ways, etc. Therefore, the numbers can be arranged in 1008! ways (cyclic permutation of 1009 elements).
2) $C=1009$. In this case, the numbers will be grouped as follows: $(1,1010),(2,1011), \ldots, (1009,2018)$. The arrangement of the numbers, as in the first case, can be done in 1008! ways.

Finally, the numbers can be arranged in $2 \cdot 1008!$ ways.