Can a collision of 2 objects result in one object having a higher velocity than either object had before they collided? Assume I had 2 perfectly elastic billiards balls of the same size and mass, that can roll on an endless billiards table losing no kinetic energy to friction.  
Is there a set of 2 velocities v1 and v2 (rate and direction) and start time and start position such that the balls collide and the result is that either one of the ball's velocities is greater than v1 and greater than v2?
Same question if the balls were differing masses?
 A: 
Is there a set of 2 velocities v1 and v2 (rate and direction) and
  start time and start position such that the balls collide and the
  result is that either one of the ball's velocities is greater than v1
  and greater than v2?

For a single pair of equal mass objects, the answer is no.  
If we look at a collision of two masses in the frame where their center of mass is at rest, each ball will depart with the same speed (or less) that it arrived with. 
If we look at the collision in any other frame, then any boost we give to one of the outgoing objects simultaneously boosts the other incoming object.  So you can't increase the speed that way.
But if the objects have unequal mass (or you have more than two so that a couple can "gang up" on another), then it's no problem.  View the collision in a frame where the center of mass is moving toward the lighter mass (as an example, one where the lighter mass is at rest).  In that frame, the lighter mass will leave with greater speed than the heavier mass arrived with.

Well what I was really wondering was can a bunch of slow moving
  billiards balls all collide such that one of them get propelled away
  at a higher velocity than any of them had had before they collided?
  I.e. can the opposite of a break shot be done?

Well, that specific example is difficult to control.  It's theoretically possible, yes.  Just dramatically unlikely.  It's just a special case of the first example where you have a heavy mass object (in this case all the balls in the rack) striking a lower mass object (the cue ball).
Something not quite the same, but similar is the case where you drop multiple balls on the ground in a vertical line.  If you arrange them by mass, then you can get the top ball to rebound much higher/faster than any of them began with.
A video of this effect is visible at Stacked Ball Drop YouTube

The resulting velocity may only be <= the closing velocity of the 2
  objects that collided.

If you can pick an arbitrary frame, then this is incorrect.  Let's expand on the train vs. tennis ball example.  In this case the train is so much more massive than the tennis ball, we can assume the center of mass to be the train.  In the frame of the train, the ball will rebound with the same speed that it impacted.
So let's imagine the train is approaching you at 30m/s and you throw a tennis ball at it at 10m/s.  This is a closing speed of 40m/s.  In the frame of the train, the ball approaches at 40m/s, then departs at 40m/s.  In your frame on the platform, the ball is thrown at 10m/s and rebounds back to you at 70m/s.  So the speed of the ball is faster (in your frame) than the closing speed of 40m/s.

BowlOfRed, if your answer were correct then I could throw a bouncy
  ball at a wall and it would deflect at a greater velocity than I've
  thrown it.

That would be difficult because most of the time you are standing still and therefore (nearly) at rest with the center of mass.  What I said earlier was that to make it happen, the center of mass had to be moving toward the collision.  If the wall isn't moving, then that condition is not true.
A: *

*First thing I want to make you consider is that Velocities are not additives. 

*When a matter collides there is a transfer of momentum collectively and not the velocity as a single component. 

*If a heavy mass strike a smaller mass then the collective momentum increases but collective velocity decreases. 

*It is evident from the Newton's momentum equation 
M1V1=M2V2

