1
2 +
1
6 +
1
12 +…
1
n(n+1) =
2003
2004 ,
1-
1
2 +
1
2 -
1
3 +
1
3 -
1
4 +…+
1
n -
1
n+1 =
2003
2004 ,
1-
1
n+1 =
2003
2004 ,
n
n+1 =
2003
2004 ,
∴n=2003.
故答案为:2003.
1
2 +
1
6 +
1
12 +…
1
n(n+1) =
2003
2004 ,
1-
1
2 +
1
2 -
1
3 +
1
3 -
1
4 +…+
1
n -
1
n+1 =
2003
2004 ,
1-
1
n+1 =
2003
2004 ,
n
n+1 =
2003
2004 ,
∴n=2003.
故答案为:2003.