Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is $$2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392$$ so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
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$\{1,2\cdots 6097392\}$ contain the integers from $1$ to $6048$ , so pair these numbers as follows:
$$1, 6048$$
$$2, 6047$$
$$3, 6046$$
$$\cdots$$
$$3024, 3025$$
When we remove any $2016$ integers from the set $\{1,2,\cdots N\}$ , clearly we can remove numbers from at most $2016$ of the $3024$ pairs above, leaving at least $1008$ complete pairs. To get a sum of $N$ , simply take these $1008$ pairs, all of which sum to $6049$ . The sum of these $2016$ elements is $1008 \cdot 6049 = 6097392$ , as desired.
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We have shown that $N$ must be at least $6097392$ , and that this value is attainable. Therefore our answer is $\boxed{6097392}$ .
The problems on this page are copyrighted by the Mathematical Association of America 's American Mathematics Competitions .