Four, (Full marks 14 points) Given that the roots of the equation $x^{2}+p x+q=0$ are 1997 and 1998, when $x$ takes the integer values $0,1,2, \cdots, 1999$, the corresponding values of the quadratic trinomial $y=x^{2}+p x+q$ are $y_{0}$. Find the number of these values that are divisible by 6.
Let $y=(x-1997)(x-1998)$ be divisible by 6. Since when $x$ takes integer values, all $y$ values can be divisible by 2, we only need to examine the cases where the factors are divisible by 3.
$$
\begin{array}{l}
\text { (1) When } x-1997=3 k \text {, then } x-1998=3 k-1 \text {. } \\
\because 0 \leqslant x \leqslant 1999, \therefore-1997 \leqslant 3 k \leqslant 2 .
\end{array}
$$

In this case, $k$ can take 666 values from -665 to 0.
（2）When $x-1998=3 k$, then $x-1997=3 k+1$.
$$
\because 0 \leqslant x \leqslant 1999, \therefore-1998 \leqslant 3 k \leqslant 1 \text {. }
$$

In this case, $k$ can take 667 values from -666 to 0.
Thus, the total number of values that can be divisible by 6 is
$$
666+667=1333 \text {. }
$$