Find all positive integers $x, y$ for which:

$$
1+x+y+xy=2008 \text{. }
$$
We will rearrange the equation to the form

$$
(1+x)(1+y)=2008
$$

and we are looking for the factorization of the number 2008 into the product of two factors greater than 1. Due to the symmetry of the problem, it is clear that it suffices to investigate only the cases where $x \leqq y$. Since $2008=2^{3} \cdot 251$ and 251 is a prime number, we can easily compile a table of all possible factorizations:

| $1+x$ | 2 | 4 | 8 |
| :---: | :---: | :---: | :---: |
| $1+y$ | 1004 | 502 | 251 |
| $x$ | 1 | 3 | 7 |
| $y$ | 1003 | 501 | 250 |

The equation has six solutions: $(1,1003),(1003,1),(3,501),(501,3),(7,250),(250,7)$.

[3 points for rearranging the left side of the equation, 1 point for finding the factorization of the number $2008$, 2 points for all solutions]