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
$$\sigma_i = (2014 + i, 1 + i, 2013 + i, 2 + i, \ldots, 1007 + i, 1006 + i), \quad 1 \le i \le 2014,$$
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
$$(2013, 2, 2011, 4, 2009, 6, \ldots, 2012, 1),$$
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 $b$.