I also wondered about this - the limitations imposed on solutions of elastic collisions by the dual laws of conservation of both momentum and kinetic energy - and whether there is there a combination of masses that can result in two masses moving with the same velocity after a collision.
Many here have quoted as presumption a priori that the situation implies an inelastic collision, but it wasn't immediately obvious to me that this must be so.
Solving the simultaneous equations got very messy very quickly, so I plotted the equations for a theoretical mass of unit one and velocity=2 colliding with a identical stationary mass in a single dimension (perhaps carriages on a rail, or perfectly straight collision of billiard balls). The plots show the results, post collision, for the velocities of mass 1 ( x axis) and mass 2 (y axis), under conservation of momentum and conservation of kinetic energy. The only possible solutions must satisfy both, thus we examine the intersections:
There are as expected two solutions, one impossible, in which mass 1 essentially passes through mass 2 unimpeded, with no effect on mass 1 (velocity mass 1 = 2, velocity mass 2 =0), and the other (and only feasible solution) with mass 1 stopping and imparting its full momentum to mass 2, which now moves away with velocity 2 (think of the white cue ball stopping dead when hitting another ball full on).
Now, if we consider the case with mass 2 twice that of mass 1:
We see the feasible solution is that in which the lighter mass 1 rebounds off heavier mass 2 (mass 1 velocity ~ -0.65, mass 2 forward at ~ 1.3). Clearly, any case in which mass 1 is lighter will see it rebounding to some degree, and mass 2 being propelled forward. So there is no possibility for $m1 < m2$ in which they can merge with the same velocity.
Looking at the opposite case, with $m1 > m2$, in this case mass 1 double that of mass 2:
We now see mass 1 retaining positive momentum (velocity ~0.65) and mass 2 shooting off rapidly (velocity ~2.7). From this, we can deduce that for any $m1 > m2$, m2 will achieve velocity >2, and the original mass 1, having imparted some momentum to mass2, will retain velocity within bounds #0< v2 < 2$. Again, there is no solution of any two masses satisfying conservation of both momentum and energy which allows the two masses to do anything other than "bounce off" each other two some degree - which is I suppose what you might infer from an elastic collision!
I haven't investigated this further, but I see no reason to suspect the result would not generalise to elastic collisions in which both masses are moving, or to to or three dimensions.
Also, I was interested to find that as ratio of $m1:m2$ was increased, the velocity achieved by mass 2 was limited to 4 (or, I suspect, double whatever the initial velocity was) - again, I suppose there has to be some limit, especially considering that at the same time (mass 1 getting relatively much heavier) mass 1 velocity approaches a limit of 2 (it is pretty much unaffected by the small mass 2) - so there is little momentum loss, and mass 2 velocity is necessarily restricted.
So I hope you can see, as I now can (and as was presumably obvious to the more experienced respondents here) there is no possibility for both masses to experience the same velocity under an elastic collision. Also, that the behaviour is fully deterministic by solution of the simultaneous equations!
EDIT I guess it makes sense that the maximum post-collision velocity for mass 2 is double that of (an relatively very heavy) mass 1 - this can be seen be adjusting the frame of reference such that mass 1 is stationary and mass 2 approaching it with velocity = -2 ..... a heavy enough mass 1 will act much like a wall, and mass 2 will bounce off it elasticly with velocity = +2, a net change of +4 .... thus the maximum velocity differential $|v2 - v1|$ is bounded by the magnitude of the initial velocity differential (as when a particle bounces off a wall).