, 2005

Solution:

The answer is 10. Adding 25 to both sides we obtain the equivalent relation $x y + 5(x + y) + 25 = 2030$ from which, collecting on the left-hand side, we get $(x + 5)(y + 5) = 2030 = 2 \cdot 5 \cdot 7 \cdot 29$. Since $x$ is an integer, then $x + 5$ is also an integer, and the same holds for $y$. Hence it suffices to distribute the prime numbers of the factorization of 2030 between the two factors $x + 5$ and $y + 5$, in all possible ways.

Since $x$ must be positive, then $x + 5$ must be at least 6, and the same for $y$. In particular, the two factors must both be positive. Moreover the equation is symmetric in $x$ and $y$, so one can assume that the unique factor 2 divides $x + 5$. For $x + 5$ there are then 5 possibilities: $2 \cdot 5$, $2 \cdot 7$, $2 \cdot 29$, $2 \cdot 5 \cdot 7$, $2 \cdot 5 \cdot 29$. The other cases are $x + 5 = 2$, to be excluded, $x + 5 = 2 \cdot 7 \cdot 29$ and $x + 5 = 2 \cdot 5 \cdot 7 \cdot 29$, to be excluded because they respectively imply $y + 5 = 5$ and $y + 5 = 1$. The cases in which 2 divides $y + 5$ are just as many, and they are all distinct from the previous ones because in 2030 there is only one factor of 2. Therefore the answer is $5 \cdot 2 = 10$.