Multiple of 2016

If $n=1009$ , let $S=\{0,1,2,\ldots,1008\}$ . Then for $a,b \in S$ and $a \neq b$ , $a+b <2016$ and $|a-b|<2016$ . This show that this set $S$ not satisfying the condition and hence $n>1009$ .

Next, we show that $n=1010$ is sufficient. As we dealing with the multiple of 2016, it is easier to consider integers from 0 to 2015 (consider the remainder if the number if divided by 2016). We partition these 2016 numbers into 1009 'boxes', as follows $(0), (1008), (1,2015), (2,2014), (3,2013), \ldots, (i, 2016-i), \ldots, (1007, 1009)$

If there is more than 1 number is selected from a particular box, then it would be $i, 2016-i$ or $i, ,i$ (note that it is congruent modulo 2016), which will satisfy the conditions.

Hence the minimum value of $n$ is 1010.

We try to make $S$ as large as possible without any two elements having a difference or sum divisible by 2016, i.e. for elements $a,b$ , $a \pm b \neq 0 \pmod{2016}$ .

So, we work on the elements of $S$ mod $2016$ . Now, if $a$ is in the set, then $b$ must NOT be in the set if $a \equiv b \pmod{2016}$ or $a \equiv 2016-b \pmod {2016}$ . It is then easy to see that we can fit at most $1009$ elements into the set, meaning $\boxed{1010}$ is the minimum number of elements required to satisfy the problem requirements.