Sum of consecutive integers is 2016

Let the first term of the consecutive integers be $a$ . Then we have $\frac{n}{2}\left(2a +n-1\right)=2016$ . Note that $2a +n-1$ is a positive integer. This means that $\frac{n}{2} \le 2016$ and hence $n\le 4032$ . Now $n=4032$ is possible, by letting $a=-2015$ .

My solution is analogous to Chan's.

The sum of $n$ consecutive integers is $n\times a_{avg}=2016$ . The minimum of $a_{avg}$ is $\frac{1}{2}$ so that the maximum of $n$ is $2\times 2016=\boxed{4032}$