Solution?

The answer seems to be 2x1008!

JBMO2020 wrote: The answer seems to be 2x2018! This is too large, there aren't even this many arrangements of the numbers without any conditions.

Sorry for the typo. It's 1008 instead of 2018

Well I think the answer is $2.1008!$ and here is (I hope) a valid solution: We look at the small cases to get a better understanding of the general problem . If we denote the number of such arrangements as $T(2k)$ for any number $n=2k$ (the problem is valid only for even numbers) we get that $T(2)=0, T(4)=0, T(6)=4, T(8)=0, T(10)=48$. Now our assumption is that if $k$ is even$T(2k)=0$ and if $k$ is odd $T(2k)=2(k-1)!$ .Now lets look at the problem from another perspective: let the points of a regular polygon with $2k$ sides have values $A_{1}, A_{2}, A_{3},\cdots A_{k+1},\cdots A_{2k}$ and this is a permutation of the set $1,2,...,2k$ , where $A_{x}$ and $A_{x+k}$ are "antipods" (indexes are $(\text{mod } 2k)$). Now lets assume $A_{1} > A_{k+1}$ (if it is not, then we can rotate the circle and rename the points as $A_{x}$ now is $A_{k+x}$ reduced $(\text{mod } 2k)$). Let $A_{1} - A_{k+1} = m > 0$. Now we know that $A_{1}+A_{2} = A_{k+1}+A_{k+2}$ so we get $A_{2} - A_{k+2} = -m$. Furthermore if $x < k+1$ then $A_{x} - A_{k+x} = ( (-1)^{x+1} )m$. Finally $A_{k}+A_{k+1}= A_{1}+A_{2k}$. Well now we get $A_{1} - A_{k+1} = ( (-1)^{k+1} )m$ so if $k$ is even then $m = -m$ so $m=0$, so $A_{1} = A_{k+1}$, which is impossible. So now we will solve for $k$ odd. We saw that we can pair the numbers $(A_{1}, A_{k+1}) , (A_{2}, A_{k+2})$ and so on with the pairs $(1, m+1), (2, m+2)\cdots (2k-m, 2k)$ in some order with the possibility that $A_{x} = q+m , A_{k+x} = q$ for some $x$ even , $x<k$ . Okay, we can now look at our problem with $n = 2018$. Lets find those $m$ that could potentially give us solutions. $A_{1}+A_{2}+\cdots +A_{2k}$ is equal to $2(A_{i}+ A_{j}+\cdots +A_{l}) + 1009m$ , because if $A_{x} + m = A_{k+x}$ then $A_{x} + A_{k+x} = 2A_{x} + m$. But $(\text{mod } 2)$ we get that $A_{1}+\cdots+A_{2k} = 1+2+3+...+2018=1009.2019$ is odd so $2(A_{i}+\cdots+A_{l}) +1009m$ must be odd, so we get that $m$ is odd. Furthermore, we know that the set $(1,2,\cdots,2018)$ can be partitioned in pairs so that the absolute value of the difference in each pair is $m$, so we get the pairs $(1,m+1), (2,m+2), .... (m, 2m)$ so we have partitioned the set $(1,2,... , 2m)$ and we get that $2m\vert 2018$ (otherwise such a partition would not be possible). Also $m$ is odd so we get $m=1$ or $m=1009$. Now we will prove that the number of arrangements for any $n=2(2k+1)$ and $m\vert 2k+1$ are $(2k)!$. We can just put down all numbers that are the small ones from their pair( for $m=1$ those are all odd numbers ) and then write the rest of the numbers inbetween them as we know that the diameter differences change as $m, -m, m, -m, \cdots$. We can do this in $(2k)!$ ways, because we can put $r+1$ numbers around a circle in $r!$ ways( imagine this as fixing $1$ and looking at the rest as a line of numbers, where the line starts at $1$ and ends at the $2k+1$-st number). Now we have our answer $T(2018) = 2.1008!$ as we have $1008!$ ways for $m=1$ and $1008!$ for $m=1009$ and we are done.