The sum of the first two numbers is 1004, so the third number must be 1005.

The sum of the last two numbers is 1005, so the middle number must be 0.

Therefore the product of all three numbers is 0.

The sum of the first two numbers and the last two numbers is 1004 + 1005 = 2009.

This counts the middle number twice.

But the sum of all three numbers is 2009, so the middle number is 0.

Hence the product of all three numbers is 0.

Alternatively: Let the three numbers be a, b and c.

We have

$a+b=1004$

$b+c=1005$

and $a+b+c=2009$

Adding the first two equations gives

$a+2b+c=2009$

and subtracting the third equationg from this gives

$b=0$

Thus the product $abc=0$.

*This problem is taken from the UKMT Mathematical Challenges.*