999

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.*