∵在等差数列中S2n-1=(2n-1)an,∴S20072007=a1004,S20052005=a1003,又∵S20072007-S20052005=2∴d=2,又由a1=-2008∴Sn=a1n+n(n-1)2d=n2-n-2008n,∴S2008=-2008故答案为:-2008