We will prove we can choose at most $\boxed{576}$ integers. First, the construction: Simply consider the sequence $1,4,8,11,15,18,22,25...,2013$ where the difference between consecutive integers alternates between $3$ and $4$. Note that the $2n$th integer in this sequence is of the form $7n-3$. Thus, $2013$ is the $576$th term, and we have shown a sequence with $576$ integers exists. Now, we will prove we can not create a sequence with more than $576$ integers. We will demonstrate a proof by contradiction. Suppose we have a sequence $a_n$ with at least $577$ integers. We divide the interval $[1,2016]$ into intervals of length $7$: $$[1,7], [8,14], [15,21], ..., [2010, 2016]$$Note that there are exactly $\frac{2016}{7}=288$ of these intervals, and every integer from $1$ to $2016$ is included. By pigeonhole, at least one of these intervals must contain $3$ integers from the sequence $a_n$. WLOG let this be the first interval $[1,7]$. It's easy to see that no matter how we choose these three integers, at least one pair of them will have a difference of $1,2,6$. Thus, we have proved that the greatest number of integers we may choose is $\boxed{576}$.