17. Positive integers $x, y$ satisfy $6 x+7 y=2012$. Let the minimum value of $x+y$ be $p$, and the maximum value be $q$, then $p+q=$
【Solution】Method one: $x+y$ reaches its maximum value when $y$ is at its minimum, and when $x$ is at its minimum, it reaches the maximum value of $y$. The minimum value of $y$ is 2, at which point $x$ is 333, $x+y=335, q=335$.
$x$ is at its minimum when it is 4, at which point $y$ is 284, $x+y=288, p=288$
$$
p+q=623
$$

Method two: $x=\frac{2012-7 y}{6}, x+y=\frac{2012-7 y}{6}+y=\frac{2012-y}{6}$, which is minimized when $y$ is maximized, and maximized when $y$ is minimized.
$y<\frac{2012}{7}$, i.e., $y \leq 287$
Since $x+y$ must be an integer, we have
$p=\frac{2012-284}{6}=288$
$q=\frac{2012-2}{6}=355$
$p+q=623$