Big Fibonacci

Answer: $999$

One unknown

The first term will be biggest when the second term is smallest. Try:

$\underline{\ \ \ n\ \ \ },\quad\underline{\ \ \ 1\ \ \ }, \quad \underline{\qquad\ }, \quad \underline{\qquad\ }, \quad \underline{2004}$

$\underline{\ \ \ n\ \ \ },\quad\underline{\ \ \ 1\ \ \ }, \quad \underline{n+1}, \quad \underline{n+2}, \quad \underbrace{\underline{2004}}_{2n+3}$

$2n+3=2004 \Rightarrow 2n=2001$ which is odd

Try:

$\underline{\ \ \ n\ \ \ },\quad\underline{\ \ \ 2\ \ \ }, \quad \underline{\qquad\ }, \quad \underline{\qquad\ }, \quad \underline{2004}$

$\underline{\ \ \ n\ \ \ },\quad\underline{\ \ \ 2\ \ \ }, \quad \underline{n+2}, \quad \underline{n+4}, \quad \underbrace{\underline{2004}}_{2n+6}$

$2n+6=2004 \Rightarrow 2n=1998\Rightarrow n=999$

Two unknowns

$\underline{\ \ \ \ a\ \ \ \ },\quad\underline{\ \ \ \ b\ \ \ \ }, \quad \underline{\ \ \qquad\ }, \quad \underline{\qquad\ \ \ }, \quad \underline{\ 2004\ }$

$\underline{\ \ \ \ a\ \ \ \ },\quad\underline{\ \ \ \ b\ \ \ \ }, \quad \underline{\ a+b\ }, \quad \underline{a+2b}, \quad \underbrace{\underline{\ 2004\ }}_{2a+3b}$

So $2a+3b = 2004$

$a$ large if $b$ small

If $b=1$ then $2a=2001$, but $a$ is an integer, so $b\not=1$.

However, if $b=2$ then $2a=1998$, so the maximum possible value of $a$ is $999$