7. Given integers $p$ and $q$ satisfy $p+q=2010$, and the quadratic equation $67 x^{2}+p x+q=0$ has two positive integer roots. Then $p=$ $\qquad$ .
7. -2278 .

Let the two positive integer roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$.
Then $x_{1}+x_{2}=-\frac{p}{67}, x_{1} x_{2}=\frac{q}{67}$.
Thus, $x_{1} x_{2}-x_{1}-x_{2}=\frac{p+q}{67}=\frac{2010}{67}=30$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=31 \\
\Rightarrow\left\{\begin{array}{l}
x_{1}-1=1, \\
x_{2}-1=31 .
\end{array}\right.
\end{array}
$$

Therefore, $x_{1}+x_{2}=34, p=-34 \times 67=-2278$.