Let the first two terms of the sequence be $a$ and $b$
respectively. Then the next three terms are $a+b$, $a+2b$ and
$2a+3b$. So $2a+3b=2004$. For $a$ to be as large as possible, we
need $b$ to be as small as possible, consistent with their both
being positive integers. If $b=1$ then $2a=2001$, but $a$ is an
integer, so $b\neq 1$.
However, if $b=2$ then $2a=1998$, so the maximum possible value of
$a$ is $999$.
This problem is taken from the UKMT Mathematical Challenges.