Just observe it

First thing we get that, $\sum_{n=1}^{n-1} f(n)=f(n)(n^2-1)..............(1)\\ \sum_{n=1}^{n-2} f(n)=f(n-1)((n-1)^2-1)..............(2)\\ \Longrightarrow (1)-(2)=f(n-1)=f(n)(n^2-1)-f(n-1)(n^2-2n)\\ \Longrightarrow f(n-1)(1+n^2-2n)=f(n)(n^2-1)\\ \Longrightarrow f(n-1)((n-1)^2)=f(n)(n-1)(n+1)\\ \Longrightarrow f(n-1)(n-1)=f(n)(n+1)\\ \Longrightarrow \dfrac{f(n)}{f(n-1)}=\dfrac{n-1}{n+1}$ .Now,you can put values and multiply and all the terms except for two will cancel.: $\dfrac{f(2)}{f(1)}\times \dfrac{f(3)}{f(2)}\times \dfrac{f(4)}{f(3)}...\times\dfrac{f(2003)}{f(2002)}\times \dfrac{f(2004)}{f(2003)}\\ =\dfrac{\dfrac{2003!}{1}}{\dfrac{2005!}{2}}\\ =\dfrac{\dfrac{1}{1}}{\dfrac{2005\times 2004}{2}} =\dfrac{1}{2005\times 1002}$ .And the $\text{L.H.S}=\dfrac{f(2004)}{f(1)}$ .Hence, $\dfrac{f(2004)}{f(1)}=\dfrac{1}{2005\times 1002}\\ \Longrightarrow \dfrac{f(2004)}{2005}=\dfrac{1}{2005\times 1002}\\ \Longrightarrow \dfrac{1}{f(2004)}=1002.$ Hence answer is $\huge{\boxed{\boxed{3}}}$ .

Induction:

$\displaystyle{f\left( 1 \right) =\cfrac { a }{ 1 } \quad ,f(2)=\cfrac { a }{ 3 } ,\quad f(3)=\cfrac { a }{ 6 } ,\quad f(4)=\cfrac { a }{ 10 } ,\quad f(5)=\cfrac { a }{ 15 } ........\\ \underbrace { f\left( 1 \right) =\cfrac { a }{ 1 } \quad ,f(2)=\cfrac { a }{ 3 } }_{ diff\quad of\quad Dr\quad =2 } ,\quad \underbrace { f(2)=\cfrac { a }{ 3 } ,\quad f(3)=\cfrac { a }{ 6 } }_{ diff\quad ofDr\quad =3 } ,\quad \underbrace { f(3)=\cfrac { a }{ 6 } ,\quad f(4)=\cfrac { a }{ 10 } }_{ diff\quad of\quad Dr=4 } ,\underbrace { f(4)=\cfrac { a }{ 10 } ,\quad f(5)=\cfrac { a }{ 15 } }_{ diff\quad of\quad Dr=5 } \quad .....}$

Clearly Denominator (Dr) terms have 2nd order difference constant , So it must satisfy 2nd order polynomial , Hence

$\therefore \quad { D }_{ n }=A{ n }^{ 2 }+Bn+c\\ \because { D }_{ 1 }=1\quad ,\quad { D }_{ 2 }=3\quad ,\quad { D }_{ 3 }=5\quad \\ \Rightarrow { D }_{ n }=n(n+1)/2\\ f(n)=\cfrac { 2a }{ { D }_{ n } } =\cfrac { 2f\left( 1 \right) }{ n(n+1) }$

Now re-checking this by induction . $for\quad n=1\quad :\quad RHS=\cfrac { 2f(1) }{ 1.2 } =f(1)=LHS$

Let f(K) is true for some K , then we have to prof f(K+1) is also true , to complete our target .

$\displaystyle{for\quad n=K\quad \quad \quad :\quad f(k)=\cfrac { 2f(1) }{ k(k+1) } \quad ....(A)\\ for\quad n=K+1\quad :\quad RHS=\cfrac { 2f(1) }{ (k+1)(K+2) } =f(K+1)=LHS\\ \\ Hence\quad \boxed { f(n)=\cfrac { 2f\left( 1 \right) }{ n(n+1) } \quad \forall n\in N } }$

With $f(1) = 2005, f(2004) = \frac{2 \times 2005}{2004\times 2005} = \frac1{1002}$ . Answer is simply $1+0+0+2=3$ .