jmc

Adding the equations, we get $$2x^2+12x-8y+26=0,$$or $x^2+6x-4y+13=0.$ We can write this equation as $$(x+3{)}^2=4(y-1).$$This is the equation of the parabola with focus $(-3,2)$ and directrix $y=0.$



By definition of a parabola, for any point $P$ on the parabola, the distance from $P$ to the focus is equal to the distance from $P$ to the $y$-axis, which is the $y$-coordinate of the point.

Subtracting the given equations, we get $2y^2-40y+118=0,$ or $y^2-20y+59=0.$ Let $y_1$ and $y_2$ be the roots of this quadratic. Then the $y$-coordinate of each point of intersection must be either $y_1$ or $y_2.$

Note that the equation $x^2+y^2+6x-24xy+72=0$ represents a circle, so it intersects the line $y=y_1$ in at most two points, and the line $y=y_2$ is at most two points. Therefore, the $y$-coordinates of the four points of intersection must be $y_1,$ $y_1,$ $y_2,$ $y_2,$ and their sum is $2y_1+2y_2.$

By Vieta's formulas, $y_1+y_2=20,$ so $2y_1+2y_2=\boxed{40}.$