This is so provided that the velocity changes continuously in which case we can apply the mean value theorem.
Velocity (v) is the derivative of displacement (x) :
v = dx/dt
Monk 1 arrives after a time t* and Monk 2 too.
Name v1(t) and v2(t) their respective velocities throughout the trajectory.
Then we know that both average velocities were equal :
avg1 = avg2
and avg = integral ( v(t) , t:0->t*) / t*
so
integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*)
which is the same of saying that the covered distances after t* seconds are the same
=> integral (v1(t) - v2(t) , t:0->t*) = 0
Thus, name v#(t) = v1(t) - v2(t) , then we obtain
=> integral ( v#(t) , t:0->t*) = 0
Name the analytical integral of v#(t) = V(t) , then we have
=> V(t*) - V(0) = 0
=> V(t*) = V(0)
So there exist a c in [0, t*] so that
V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem)
We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get
V'(c) = v#(c) = v1(c) - v2(c) = 0
=> v1(c) = v2(c)
So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2.