Iranian Mathematical Olympiad

There are $2014\times 2013$ ordered pairs $(a,b)$ such that $1\le a\ne b\le 2014$. On the other hand, there are $2013$ consecutive pairs in each permutation. Therefore, the number of such permutations is at most $2014$.

Consider the $2014$ permutations of the form $${\sigma }_i=(2014+i,1+i,2013+i,2+i,\dots ,1007+i,1006+i),1\le i\le 2014,$$ where all the numbers are considered modulo $2014$ ($2014$ itself being an exception). We claim that for each $a\ne b$ there is exactly one index $i$ such that $b$ has appeared after $a$ in ${\sigma }_i$. To prove it, note that the differences of consecutive terms in all these permutations are $$(2013,2,2011,4,2009,6,\dots ,2012,1),$$ and all the numbers are modulo $2014$. So the position of $b$ and $a$ is uniquely determined by the value of $b-a$ and the index of permutation is uniquely determined by the value of $b$.