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 $|I_\sigma| \neq 2014$. 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 $|I_\sigma| \neq 2014$ and $|I_\sigma|$ takes values $1, 2, \dots, 2013$ only.