Let the first two terms of the sequence be $a$ and $b$ respectively.

Then the next three terms are $a+b$, $a+2b$, $2a+3b$. So $2a+3b = 2004$.

For $a$ to be as large as possible, we need $b$ to be as small as possible, consistent with both being positive integers.

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 $999$, giving us the first five terms:

999, 2, 1001, 1003, 2004

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.