Integral Solutions

Since $2003$ is prime, its only factors are $\pm 1$ and $\pm 2003$ . Since $xyz$ has to equal positive $2003$ , therefore either all three variables can be positive, or two can be negative and the other positive. That gives us three cases of possible factors: $-1,-1,2003$

$1,1,2003$

$1,-1,-2003$

For the first two cases, there are $3$ ways to "choose" which variable is which factor, since two of the factors are the same. And for the last one, there are $3!$ or $6$ ways to choose. Therefore, $3+3+6=\boxed{12}$

Since 2003 is checked to be a prime number,

1 * 1 * 2003

1 * 2003 * 1

2003 * 1 * 1

To add in two negative's for each of them, each would have 3 C 2 ways to add with the positive to 4 cases.

4 x 3 = 12

Answer: 12