Junior Mathematical Olympiad

Let us consider a configuration satisfying the conditions of the problem. Let $A,B$ be two consecutive numbers in that configuration and let their diametral opposite numbers be $a,b$, respectively. Then holds $A+B=a+b$ i.e. $A-a=b-B$. Since $A,B$ are arbitrary, we get that every difference of the numbers which are on the diametral opposite vertices is constant. That means $A-a=C$, where $C$ is a constant. Hence, the numbers from $1$ to $2018$ should be paired in $1009$ pairs such that the difference of the numbers in every pair is $C$. Such pair can be formed if and only if $C$ is a divisor of $1009$.

Let $C=k$. Then it must (respectively) the numbers to be paired on the following way $$\begin{gathered} \{1,2,\dots ,k\}\to \{k+1,k+2,\dots ,2k\},\{2k+1,2k+2,\dots ,3k\}\to \{3k+1,3k+2,\dots ,4k\},\dots , \\ \{2019-2k,\dots ,2017-k,2018-k\}\to \{2019-k,\dots ,2017,2018\}\dots \dots (1) \end{gathered}$$

($k$ elements of one set are paired with $k$ elements of the other set, such that every element of the set $\{1,2,\dots ,2018\}$ should appear only once). If the number of (arrows) pairings in (1) is $m$, then $2018=2km$ i.e. $1009=km$. Since $1009$ is prime, the only possible values for $k$ are $1$ and $1009$.

1) $C=1$. In this case the pairing of the numbers will be $(1,2),(3,4),\dots ,(2017,2018)$ and the numbers in the pair will be diametral opposite. Let us fix on one diameter the first pair $(1,2)$. Let us note that it is necessary to choose the configuration between $1$ and $2$ in one direction (for example clockwise), and the other half of the numbers will be determined by the condition of the problem. Since we do not count the rotations, the next number to the number $1$ (clockwise) can be chosen in $1008$ ways, and then next number of the chosen one in $1007$ ways etc. So, the number of all configurations in this case is $1008!$ ways (cyclic permutations on $1009$ elements).

2) $C=1009$. In this case the numbers will be paired in the following way $(1,1010),(2,1011),\dots ,(1009,2018)$. Therefore, the discussion about the configurations is the same like in the previous case, so there are $1008!$ configurations in this case.

Finally, the number of all configurations is $2\cdot 1008!$