$n$ is clearly odd, so we can write the series as:
$1+(-2+3)+(-4+5)+...+(-(n-1)+n)$. So the sum is the same as $1+1+...+1=(n+1)/2$.
So $(n+1)=4016$, so $n=4015$
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
This problem is taken from the UKMT Mathematical Challenges.