Can a virtual image act as a virtual object for another optical device? We call an image virtual if it is made by intersection of extension of diverging rays. And real if it is made by intersection of converging rays.
Assume a real image is made somewhere. Now if we place an optical device before the intersection point, that image now is like a virtual object for the new device.
We can say a virtual object is like a defeated real image. Rays weren't able to intersect normally, instead, they go through another device.
Now, assume a virtual image. Can this kind of image act as a virtual image for another optical device? 
 A: I am doing just what the OP wonders as I type this.
The lens inside my eye is exactly the correct power to take an object at infinity and form a perfect real, inverted image on the retina of my eye.  So I can drive without glasses, and I could answer SE questions using a giant video screen.
Unfortunately, my monitor is small, only about 1 metre away, and quite out of focus.
So I place a converging lens a few inches in front of my eye.  This lens forms a virtual image of my monitor at infinity, so the lens of my eye can focus it on my retina.  The virtual image of the screen serves as a real object for the lens in my eye.
Consider a person who has surgery to correct for cataracts, and through some strange screw-up has a diverging lens inserted in their eye.
To correct this error, the victim would use a very powerful converging lens in their glasses.  This would form a real inverted image somewhere in the victim's eye, between the lens of the eye and the retina.
From a ray point of few, the diverging lens would "unbend" these rays just enough to move the image to the retina.
From a lens formula point of view, the real image formed by the glasses would be a virtual object for the eye lens.  The sign convention for the particular version of the lens formula would apply, and the result would be a real inverted image on the retina...
