In the end, we have to throw up our
hands and say there isn't an "answer" answer. Some will be
tempted to say "but wait, this could really happen! There
has to
be a final outcome!"
Sorry, that's wrong. We have idealized our experiment
to a great extent:

The bicycles and the fly all start at the same point.

The bicycles travel at a constant speed and never waver.

The fly travels at a constant speed, and never wavers. Not only that,
there is no turnaround time between trips.
If we're going to accept all of these idealized conditions, we have to
accept an idealized set of equations to solve it, with a
singularity
just where we were expecting an answer. So:
Mu.
And don't ask me if the fly has a
buddhanature or I will knock you into
the river.
Of course, anything worth doing is worth expending far too much effort
on, and making yourself a fool over. Here's the rest of my original writeup, just so you know
what
ariels is on about.
First of all, let's dispense with all of these silly "
kilometer"
thingies. I will measure everything in a distance unit of my
own invention, the "mu", equivalent to 10 km. This simplifies
our thought experiment a bit:

The track extends from 1 mu to +1 mu.

The bicycles travel at 1 mu/hr.

The fly travels at 2 mu/hr.
At the beginning of the experiment, the track looks like this:
+oo*oo+
1
0 1
So let's try ariels' suggestion and pick a small time t
that the fly waits before starting its flying. In our system, time is
equivalent to position, so we can also say that the fly starts at a positive
position d=2t at time 0. Now let's assume that the
fly passes the bicycle traveling to the right and flies all the way to
the bicycle traveling to the left (well, we can say the fly alighted on
the righttraveling bicycle and began its next trip, so we aren't really
assuming anything).
*
<oo+oo>
d d
The fly starts out d from the lefttraveling bicycle,
and catches it at d after an elapsed time of d,
traveling 2d.
*
<oo+oo>
d d
For the second trip, the fly is 2d from the righttraveling
bicycle, and catches it at +3d after an other 2d,
traveling 4d.
For the third trip, the fly is 6d from the lefttraveling
bicycle, and catches it at 9d after an elapsed time of
6d,
traveling 12d.
For the fourth trip, the fly is 18d from the righttraveling
bicycle, and catches it at +27d after an elapsed time of
18d,
traveling 36d.
For a particular d, we add up the trips 2d + 4d + 12d + ... +
4*3^{t2}^{ }and find where the next trip would
exceed 2 mu, the remainder giving us our final position.
We have a problem! If you try different values for d,
you will see that for a d=3^{2n }the fly will
end up at one bicycle, for d=3^{2n+1 }the fly
will end up at the other bicycle! Not only that for d=3^{2n}/2,
the fly will end up in the middle! As d gets
closer and closer to 0, the ending point travels continuously
back and forth across the entire track from 1 to +1 and back again.
A particularly strange attractor which fills our entire problem space!
Ok, so let's
work the problem backwards: We need to find the
final position, where if you backtrack to the beginning, the fly will have
traveled 2 mu. We pick a final position and measure the distance
d to the bicycle the fly left on the last trip. We can see right off that
no matter where the fly ends up, the last trip will be 2/3*d mu, and will
have taken 1/3*d hr.
So if the fly ends up at one of the bicycles, all the way at 1,
oo*+oo
1
0 1
the previous trip will have begun 1/3 of the way into the trip, at +1/3.
The fly travels 4/3 during this trip, and is 2/3 from the other bicycle:
+oo+*oo+
1
0 1
The trip before that will have begun 1/9 of the way into the trip, at
1/9. The fly travels 4/9 during this trip, and is 2/9 from the other
bicycle.
+oo*+oo+
1
0 1
So each trip will begin at 3^{t} (t being the number of trips
until the end), will take 2/3^{t} hr, and causes the fly to travel
4/3^{t} mu.
Thus:
4/3 + 4/9 + 4/27 + 4/81 + .... = 2.
So the fly could end up at 1. But there's a problem!
Assume the fly ends up in the middle:
oo+*+oo
1
0 1
The previous trip will have begun 2/3 of the way into the trip, at +2/3.
The fly travels 2/3 during this trip, and is 4/3 from the other bicycle:
+oo+*oo+
1
0 1
The trip before that will have begun 2/9 of the way into the trip, at
2/9. The fly travels 8/9 during this trip, and is 4/9 from the other
bicycle.
+oo*+oo+
1
0 1
So, the fly travels 2/3 + 8/9 + 8/27 + 8/81 + 8/243 + ... = 2.
Oh crap. In fact, no matter where the fly ends up, it has traveled
exactly 2 mu since the beginning. Well, we probably could have predicted
that from our previous results.