Good a time as any to burn my first post, I guess...
The spirit of the problem as stated is that the speed of the conveyer is always equal and opposite the speed of the airplane relative to the conveyer surface, or the speed of the airplane's wheels. If this is true, then by definition the airplane's speed through the air is zero. Thus, no takeoff.
However...there is a problem:
Start with F = ma.
There are two forces acting on the aircraft: a thrust force and a frictional force. The frictional force is simply some dimensionless coefficient c multiplied by a Normal force
N=mg, where g is the gravity constant.
So far we have T - cmg = ma
=> a = (T-cmg) / m.
T, c, m and g are all constant, therefore, a is constant.
First, lets look at an airplane on a normal runway. The velocity of the wheels, Vw, is simply equal to acceleration multiplied by time:
Vw = at
Introduce an opposite moving conveyer belt velocity, Vc, and the equation becomes
Vw = at + Vc
In the problem statement, Vw = Vc, but this would lead to:
Vc = at + Vc
Which, after t > 0, is mathematically impossible unless
a = 0
=> T = cmg
which we know is not true (if it were, then the plane would not be able to take off at all).
So I guess the real answer to the original problem is that there is no answer; the conditions imposed are impossible, not just practically, but theoretically.