Saudi Arabia Mathematical Competitions

The greatest element of set $A_n$ is $$1+2+\dots +n=\frac{n(n+1)}{2}$$ and the smallest element of $A_n$ is $$-1-2-\dots -n=-\frac{n(n+1)}{2}.$$ Also, the difference of any two elements of $A_n$ is even, hence all elements of $A_n$ are of the same parity. Let us prove that all integers between $-\frac{n(n+1)}{2}$ and $\frac{n(n+1)}{2}$, and of the same parity with $\frac{n(n+1)}{2}$, belong to $A_n$, and these are all the elements of $A_n$. Indeed, let $x\in A_n$ be an element such that $-\frac{n(n+1)}{2}\le x<\frac{n(n+1)}{2}$. Case 1. If the writing of $x$ begins with -1 , then by changing -1 by +1 , we get $x+2\in A_n$. Case 2. If the writing of $x$ begins with +1 , then consider the first term with sign -. A such term exists, otherwise $$x=1+2+\dots +n=\frac{n(n+1)}{2}.$$ In this case we have $$x=+1+2+\dots +(k-1)-k\pm \dots \pm n,$$ where $-k$ is the considered term. By changing the signs of terms $k-1$ and $k$, it follows $x+2\in A_n$. We obtain $A_n=\left\{-\frac{n(n+1)}{2},-\frac{n(n+1)}{2}+2,\dots ,\frac{n(n+1)}{2}-2,\frac{n(n+1)}{2}\right\}$, that is the number of elements in $A_n$ is $\frac{n(n+1)}{2}+1$.