If σ={2014,2013,…,1008,1,1007,1006,…,2} then ∣Iσ∣=2013.
If σ={2013,2012,…,1008,1007,1,1006,1005,…,2,2014} then ∣Iσ∣=2012.
If σ={2k,2k−1,…,k+1,1,k,k−1,…,2,2k+1,2k+2,…,2014} then ∣Iσ∣=2k−1; k=1,…,1007.
If σ={2k+1,2k,…,k+2,1,k+1,k,…,2,2k+2,2k+3,…,2014} then ∣Iσ∣=2k, k=1,…,1006.
If σ={1,2,…,2014} then ∣Iσ∣=1.
In other words, ∣Iσ∣ takes values 1,2,…,2013.
Now let's prove that ∣Iσ∣=2014. If ∣Iσ∣=2014 then Iσ={0,1,2,…,2013}. Since Iσ={∣σi−i∣:i∈I}, we get ∑i∈I(σi−i)=0. On the other hand, among the numbers ±0,±1,±2,…,±2013 there is no number equal to 0 because there are 1007 odd numbers, namely 1,3,…,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 ∣Iσ∣=2014 and ∣Iσ∣ takes values 1,2,…,2013 only.