Why do two bodies affect each other with the same but opposite torque? Update: After further reading, it looks like the torque $C_{12} = C_{21}$ should not be seen as caused by the force $F_{12} = -F_{21}$. The resulting force at the contact point causes an independent torque not to be confused with $C_{12} = C_{21}$. So what force(s) then causes the torque $C_{12} = C_{21}$ ? And why are this torque the same but opposite direction for the bodies?
Original post:
Two solid (undeformable) bodies of body 1 and body 2 are in contact with each other. Let's call the body's common point of contact K.
According to the law of action and counteractivity, also known as Newton's third law, the bodies interact with the same but directive resulting force: $$F_{12} = -F_{21}$$ where $F_{12}$ is the resulting force that body 1 experiences from body 2 and vice versa at the point of contact K.
The force $F_{12}$ causes (?) A resulting torque $C_{12}$ on body 1.
The force $F_{21}$ causes (?) A resulting torque $C_{21}$ on body 2.
According to the law of action and counteractivity, $C_{12} = -C_{21}$. But why?
Or have I interpreted the law wrong, maybe it is not the force $F_{12}$ and $F_{21}$ at the contact point that give rise to the resulting torque for body 1 and body 2?  (I strongly believe that I misunderstand the connection between the resulting torque and the forces $F_{12}$ and $F_{21}$, as the following example shows):
My counter example: Ponder that a bowling ball and a ladder travel in the air and suddenly collide in a point that is furthest down on the steps. According to Newton's third law, the bodies affect each other with an equal but counter-directed force, $F_{12} = -F_{21}$.
The resulting torque for the ladder can be calculated by multiplying the force by the perpendicular distance between the force and the center of the mass, and the same for the ball.
However, the perpendicular distance from the force to the center of mass on the ladder (we assume the center of mass lies at the center of the ladder) is much greater than the bowling ball, so the ladder should experience a greater torque than the ball.
Mathematical: Torque_ladder = Force * (Distance to center of mass of ladder)> Force * (Distance to center of mass of ball).
So I have proven that the ladder experience a greater torque than the ball. Or simply expressed, the ladder wants to rotate around its center of mass more than the ball. This proof is based on assuming that the resulting torque that the ladder experience from the ball can be determined by the force $F_{12} = -F_{21}$ at their contact point and the distance from the force to their respective center of mass.
 A: I've replaced my comments by this answer,
which I hope summarizes things in a way which will be more useful to people who come across the question,
as well as conforming better with SE practice.
Much of the content of the original post is correct, 
but in general $C_{12}\neq \pm C_{21}$
or rather $\mathbf{C}_{12}\neq \pm \mathbf{C}_{21}$ (vectors). 
The diagram below illustrates
a collision between two bodies, 1 and 2.
Equal and opposite forces act at contact point K, position $\mathbf{r}_K$:
$\mathbf{F}_{12}$, on body 1 due to 2,
and $\mathbf{F}_{21}=-\mathbf{F}_{12}$ (Newton's third law), on body 2 due to 1.
Knowledge of the geometry, that is the centres of mass $\mathbf{r}_1$, $\mathbf{r}_2$, 
and the contact point $\mathbf{r}_K$, 
and the (equal and opposite) forces,
is sufficient to determine what happens in the collision.

It is often convenient to express the effects on body 1
as a combination of 
(a) a force $\mathbf{F}_{12}$ acting at the centre of mass $\mathbf{r}_1$ and 
(b) a torque $\mathbf{C}_{12}$ about an axis going through $\mathbf{r}_1$. 
In that case $\mathbf{C}_{12}=(\mathbf{r}_K-\mathbf{r}_1)\times \mathbf{F}_{12}$ 
where $\times$ is the vector cross product.
Similarly the effects on body 2 may be represented as
(a) a force $\mathbf{F}_{21}$ acting at the centre of mass $\mathbf{r}_2$
and
(b) a torque
$\mathbf{C}_{21}=(\mathbf{r}_K-\mathbf{r}_2)\times \mathbf{F}_{21}$
about an axis going through $\mathbf{r}_2$.
$\mathbf{F}_{12}$ and $\mathbf{F}_{21}$ are the same as before,
and $\mathbf{F}_{21}=-\mathbf{F}_{12}$,
but this does not lead in general to a simple relation between $\mathbf{C}_{12}$ and $\mathbf{C}_{21}$.
Instead one gets (combining the two expressions)
$$
\mathbf{C}_{12} + \mathbf{C}_{21} + (\mathbf{r}_1-\mathbf{r}_2)\times \mathbf{F}_{12} = 0
$$
which is an expression of angular momentum conservation (no overall external torque on the system).
If everything lies in a plane,
these expressions may be simplified, written in terms of perpendicular distances etc.
In the diagram below,
the torques are illustrated schematically:
really they should be thought of as vectors perpendicular to the plane of the drawing.

Knowledge of $\mathbf{r}_1$, $\mathbf{r}_2$, the (equal and opposite) forces, and the torques,
is sufficient to determine what happens.
Now it is not necessary to know the contact point $\mathbf{r}_K$.
Normally, we would actually know $\mathbf{r}_K$,
but it is often convenient to express the dynamics of each particle 
in terms of the translation of the centre of mass and rotation about the centre of mass,
so this way of expressing things is a natural one.
If one attempts to calculate the torques due to the two forces $\mathbf{F}_{12}$ and $\mathbf{F}_{21}$
about the same fixed point,
then of course one gets equal and opposite cancelling terms.
An interaction between two bodies cannot produce a net external torque acting on the combined body.
But this is not the physical situation set out in the original question.
In actual collisions between bodies,
the forces are often taken to act impulsively (i.e. instantaneously)
and their effects on the velocities and angular velocities of the two particles depend on various
physical parameters (smoothness of surfaces, moments of inertia, elastic or inelastic nature of the collision)
as well as the relevant conservation laws:
but none of these details are relevant to the original question.
