$n$ 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.

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