Fix a positive integer $n$. A permutation $(x_1,x_2,\ldots,x_{2n})$ of the set $\{1,2,\ldots,2n\}$ satisfies the property $A$ iff for at least one $i$ in $\{1,2,\ldots,2n-1\}$ , $|x_i - x_{i+1}|=n$.
I.e. If the difference between adjacent numbers in a permutation is equal to $n$, then this permutation has the property in the $i^\text{th}$ position.
Example: For $n = 3$, $(1,5,2,3,4,6)$ has the property (because $x_2=5$ and $x_3 = 2$ are next to each other and $|x_2 - x_3| = |5 - 2| = 3$); similarly $(6,3,1,2,4,5)$ has the property. However, $(1,2,3,4,5,6)$ does not have the property.
I need to prove the following statement: There are at least as many permutations with the property than permutations without it.
Can someone please point me towards how to go about this?
I tried this for small cases and can see how it is true.
I had the idea to first define an algorithm which generates all the possible permutations. Then show that the transposition from one permutation to another produces elements which are adjacent to each other and the difference between them is $n$.
The question suggests to show a bijection between the permutations that do not have this property to the permutations that do have this property, for exactly one $i\geq 2$.