There is a theoretical situation where is an infinitely wide(means
tension is equal from all directions) pool filled with a liquid. There
would be a layer of elastic material on the surface of the liquid,
such as balloon rubber. Two lead balls would be placed on the layer
with a great distance to each other. Lead balls weights would be
enough to push downwards at least so much that the center of the ball
is below the mean altitude of the layer. What would happen or could be
predicted to happen in this situation?

This sounds like something to do with gravity?

Hi,
  
   You need to clarify what you mean by "mean" level of the membrane
layer. I think I understand what you intend; that the "mean" is the
level of the membrane flush against the surface of the liquid, with no
orthogonal forces (such as the weights of the lead spheres) to distort
the membrane out of flatness. Yes?

   Otherwise, the statistical mean of the depth of the membrane would,
technically, be a function of both the undistorted depth and the
depths of the weight-distorted regions. This mean would be between the
two extremes, not on the flat surface.

   Also, by "great distance", should one assume you mean "great enough
to be practically of no effect to each other"? And additionally,
should the depth of the liquid be assumed limitless, but with a
perfectly uniform acceleration due to gravity for all depths?

   But in principle, this ought to be a theoretically tractible
problem, as long as all the relevant variables are clarified, and as
long as the question you ask of it is also clear.

I believe that the answer also depends on the specific weight of the
liquid and the elasticity of the membrane.  If the latter is extremely
elastic, it wouldn't distribute the weight of the balls to as great a
surface area, allowing them to sink deeper.

If the liquid were mercury, the balls would float without any membrane.
But maybe I am getting too complicated.

No, you're right, these are also relevant variables, since the
question of exactly what will occur in the system also depends on
these things. Nice going.

Is the ball/membrane interface frictionless? That's also important. If
everything else is smoke and mirriors that's the only thing keeping
the balls apart. The rest state is both balls together at the same
spot. That's a gut feeling and the best i can do to support it is that
the liquid displaced by the cone made by both balls will be less than
the liquid displaced by each ball separately because the slope of the
edges of the cone will be much steaper.

Man i wish there was an edit... Of course the balls with displace the
same amount of liquid together or apart. It's just that they can sit
lower when together.

I think the liquid is just a red herring.  It doesn't matter at all. 
Just think of an infinite elastic memebrane, with nothing above it or
below it.  This makes it pretty clear that the balls will roll toward
each other.

I considered that, too, but if the surface is infinite and the balls
are placed a great distance from each other, I don't think so.
Also, because the surface is infinite, the indentations in the
membrane caused by the balls will have no effect on the overall water
level, which I presume to be the "mean".
With my suggestion that the elasticity of the membrane is a factor, I
considered the ultimate situation:  the membrane so elastic that it
let the balls sink deeper, the preasure of the liquid forcing the
membrane to close around them so that they continued to sink, pulling
after them an ever lengthening collapsed tube of membrane.
Of course this is purely theoretical, but so is an infinitely large pond.

Simple, all you have to do is consider the forces involved in the situation.

Your forces up are the tension of the surface.  Your force down is g
on the two balls.  Though the force of the tension is stronger than
the force of gravity on the two balls, the force of gravity still has
some effect, and causes a depression because the force is strongest at
the center of gravity of the balls and weakens as you get further away
from each ball.  However, in this perfect situation, there is no point
where the downward force will not affect the curvature of the surface,
meaning the curvature will create a horizontal force towards the
center of mass of the two balls.  This will result in both balls
eventually colliding.

The only unanswered factor in this situation is the force of friction.
 If the force of friction is greater than the horizontal force, the
balls would never roll.  It appears, however, that the response your
instructor was looking for was that the balls would collide.

(for this problem, I assumed the surface was perfectly flat before the
balls were placed on it, which appeared to be the intent)

By the way, the use of the liquid ensures a flat surface, because
gravity would make it this way (unless another force is pushing it
upward, such as a jet of some sort, which is unlikely).  The idea of
"mean altitude" comes into effect after the balls were placed on the
surface (because, after this point, the balls have distorted the
surface to the point where it is no longer perfectly flat).

Zork,
The liquid is also providing an upward force.  As I understood the
definition of the problem, the membrane is there to distribute the the
weight of the balls, increasing displacement, but not by itself
supporting them.  If the balls were near enough to each other so that
the area of the membrane depressed by each adjoined, then, I agree,
the balls would roll together (ignoring friction).
But since the pond is infinite in size, I expect that the definition
intends that the balls are not so close.
Myoarin

Yes, it is correct that the water is creating an upward force. 
However, this force is obviously not strong enough to completely
counteract gravity, as seen by the depression created by the balls. 
Also, since the tension is perfectly equal and the plane is perfectly
flat, the radius of the depression is, in fact, infinite, though the
slope tends to become miniscule as x (horizontal distance from ball)
approaches infinity.  There is no point that is not effected by the
ball.  Hence the problem depends solely on the frictional force verses
the downward force created by the slope from the balls.

Zork,
That sounds theoretically correct, but the question states an "elastic
membrane", so I don't think that the radius of depression will be
infinite.
But I don't reckon that we are going to be able to settle the question.

Cheers, Myoarin