Torques acting on a unicyclist When a unicyclist rounds a bend, he or she has to tilt in order to generate a frictional force, the friction acting as the centripetal force constraining the unicyclist to a circular path. However, this tilting causes the combined weight of the unicyclist and the unicycle to have a torque about the point of contact of the tyre of the unicycle and the ground. What balances this torque? It cannot be the frictional force nor can it be the normal reaction on the tyre since both forces act on the point of contact and thus have no torque about it. On the other hand, if the torque remains unbalanced, the unicyclist will eventually crash to the floor yet this doesn't happen in many cases.
I am assuming the motion is observed by a non-accelerating observer, so there is no centrifugal force or any other fictitious force.
 A: This does not have an easy answer (unless you are good with mechanics).
If the unicycle were not moving, you are right, it would topple.
But because the unicycle is moving, the torque will not make it fall but rather precess. You have seen this happening multiple times: it is the same happening to a rotating gyroscope (a spinning top): if the spinning top is spinning it will not fall even if it is slightly tilted, instead it will start precessing around a vertical axis.
You ask for what torque is responsible for this and the answer is ... none and all.
Rigid bodies do not behave exactly as point masses and their equations of motions are... weird. A torque is not directly proportional to the (angular) acceleration of the body, there are extra terms appearing if the body is moving/rotating. This is because for a rigid body the mass distribution is not concentrated in a point and this leads to the forces acting differently at each point of the body. This is a big problem when you try to describe your system's rotations and torques as each point is rotating weirdly - unless you put yourself in the center-of-mass system and then things are easier [see below].
Another way to see this, is conservation of angular momentum. Because the unicycle's wheel is spinning, it has some angular momentum $L$. When gravity starts acting on it (being an external force) it will modify it (again, according to the equation of change of angular momentum, see below) in a direction which is perpendicular to both the angular momentum vector and the torque: the result is that the unicycle will round on the bend instead of falling!
You can see here for a derivation in the case of a gyroscope and here wikipedia explains the equation of motion for a rigid body.
A trick is to go the center of mass system. Here the equations of motion are simply $$\vec{\tau} = I \vec{\alpha}$$ with $\vec{\tau}$ the torque and $\vec{\alpha}$ the angular acceleration. The only torque acting is due to the reaction force $\vec{N}$ which we unfortunately do not know. We do however know its direction: opposite to the velocity of the unicycle to react to it. Because by tilting towards the curve you are not slipping, it means that the reaction force is opposing slippage: so there is a "hidden" force present that is directed towards the center of the curve and (if you use your right-hand-rule) you will see that this generates a torque that rotates the unicycle along the curve. Unfortunately, we can't compute this force easily. To solve this problem, as you did, the solution is using the contact point as a pivot. The price we pay is having to deal with extra terms in our equation of motion - but things are super easy if we describe instead the system from an angular momentum $\vec{L}$ point of view [see linked resources] because the equations simplify to
$${d\vec{L} \over dt} = \vec{\tau}$$ (although you need to be careful in computing $\vec{L}$ if your center of mass is moving and your a not in the CoM system!) and so you see that the effect of gravity (helped by the reaction force) is to change the angular momentum towards.. going around the curve.
As a final remark: even if we hide them sometimes using some specific application points as pivots, when a rigid body moves that is due to the external forces, in this case gravity and the reaction force (hidden if you write the torque equation, but present "in real life"). Torques are a way of describing rotations, but the fundamental element is always... force. And they don't disappear depending on pivotal points!
So even if it does not exert a torque (or rather, it does, but it vanishes when you choose the contact point as your pivot) the reaction force is actually there and it is helping you to bend.
A: 
On the other hand, if the torque remains unbalanced, the unicyclist will eventually crash to the floor

Not quite.  An unbalanced torque must mean that angular momentum is changing.  But angular momentum can be expressed in many different ways, not just rotation.
In the case of the unicyclist if we look at the point on the ground that the cycle passes over in a turn, then we demand that the gravity modify the angular momentum about that point.  This happens by the cyclist accelerating away.  Linear velocity (if not in line) can contribute to angular momentum.
Another way of looking at this is in the frame of the cyclist.  Here the cyclist appears to maintain the turn with no counter-torques.  But since the cyclist is circling, this is not an inertial frame.  There will be a fictional force appear in this frame.  The force will exactly counter the torque from gravity.
A: Why do you need to lean forward when standing on a train when it is accelerating? You displace your CG to be out of line with the normal force on your feet. In that case, the torque about your CG due to the normal force balances that due to the static friction on your feet.
The case of the unicyclist may be similar, though things are a little more complicated with the angular momentum of the wheel.
