If the roots of $x^2+px+q=0$ are integers, then the quadratic can be expressed as $(x+a)(x+b)$, where a and b are integers. Therefore, $a+b=p$ and $ab=q$. Add these equations up to get: $ab+a+b=p+q\\ ab+a+b=1998 \\ (a+1)(b+1)=1999$ Luckily 1999 is prime, so no factoring is needed. There are 4 solutions for a and b in the form of (a, b), which are $(0, 1998), (1998, 0), (-2, -2000), (-2000, -2)$ Plug that in for the previous $a+b=p$ and $ab=q$ to get that $p = 0+1998 = 1998, q = 0\cdot1998=0$ and $p=-2-2000=-2002, q = -2\cdot-2000 = 4000$ Therefore, the solutions are: $\boxed{(1998, 0), (-2002, 4000)}$