Ta có: 1+1/2+...+1/2007+1/2008
= (1+1/2008)+(1/2+1/2007)+....(1/1004+1/1005)
= 2009/2008+2009/4014+...+ 2009/1009020
= 2009.(1/2008+1/4014+....+1/1009020)
Như vậy, ta có: A = 1.2.3....2007.2008.(1+1/2+...+1/2007+1/2008)
⇒ A = 1.2.3...2007.2008. 2009.(1/2008+1/4014+....+1/1009020)
Do A có chứa thừa số là 2009 ⇒ A chia hết 2009 (đpcm)