Mongolian Mathematical Olympiad

If $\sigma =\{2014,2013,\dots ,1008,1,1007,1006,\dots ,2\}$ then $|I_{\sigma }|=2013$. If $\sigma =\{2013,2012,\dots ,1008,1007,1,1006,1005,\dots ,2,2014\}$ then $|I_{\sigma }|=2012$. If $\sigma =\{2k,2k-1,\dots ,k+1,1,k,k-1,\dots ,2,2k+1,2k+2,\dots ,2014\}$ then $|I_{\sigma }|=2k-1$; $k=1,\dots ,1007$. If $\sigma =\{2k+1,2k,\dots ,k+2,1,k+1,k,\dots ,2,2k+2,2k+3,\dots ,2014\}$ then $|I_{\sigma }|=2k$, $k=1,\dots ,1006$. If $\sigma =\{1,2,\dots ,2014\}$ then $|I_{\sigma }|=1$. In other words, $|I_{\sigma }|$ takes values $1,2,\dots ,2013$.

Now let's prove that $|I_{\sigma }|\ne 2014$. If $|I_{\sigma }|=2014$ then $I_{\sigma }=\{0,1,2,\dots ,2013\}$. Since $I_{\sigma }=\{|{\sigma }_i-i|:i\in I\}$, we get ${\sum }_{i\in I}({\sigma }_i-i)=0$. On the other hand, among the numbers $\pm 0,\pm 1,\pm 2,\dots ,\pm 2013$ there is no number equal to $0$ because there are $1007$ odd numbers, namely $1,3,\dots ,2013$. The result of addition and subtraction of these $1007$ numbers is an odd number, and the result of addition or subtraction of odd and even numbers is odd too. Therefore, we conclude that $|I_{\sigma }|\ne 2014$ and $|I_{\sigma }|$ takes values $1,2,\dots ,2013$ only.