Two rotating discs with same angular momentum when brought in contact completely stop. Why is the angular momentum not conserved in this case? Two discs mounted on different thin, lightweight rods oriented through their centres are made to rotate about their axes seperately such that the angular momentum of the two about their respective axes is same in both magnitude and direction. When both are brought in contact, they stop due to force of friction. Why is the angular momentum not conserved about their axes in this case?(Was positive before contact, but zero after the process completes, no external force present either)
 A: Let us start by assuming that the two discs have equal Angular momentum about their own axes which is equal to $\vec{L}$. And let the two discs be rotating with angular velocities $\vec{\omega_1}$ and $\vec{\omega_2}$ of radii $R_1$ and $R_2$ with masses $M_1$ and $M_2$ respectively (for discs 1 and 2). Let their moment of inertia be $I_1$ and $I_2$ about their respective center of masses. Thus,
$$I_1 = \frac{M_1R_1^2}{2}$$ and
$$I_2 = \frac{M_2R_2^2}{2}$$ and also,
$$\vec{L} = I_1\vec{\omega_1} = I_2\vec{\omega_2} ...(i)$$
Now, coming to the question, angular momentum is always dependent on the axis chosen. So if we consider initial and final angular momentum about axis through any of the centers then angular momentum would not be conserved by any chance as an external torque would act due to friction at their common point of contact and the distance between the axis and their common point of contact will not be zero. 
But if we consider angular momentum about the axis through their point of contact and perpendicular to the plane of discs then it must be conserved as the torque would be zero as distance between the point of action friction and the our chosen axis will be zero.
We can verify this by solving.
From equation (i) it can be seen that the angular velocity and the angular momentum will be in the same direction always.
Now considering initial angular momentum about the axis through point of contact, we will get,
$$\vec{L_{net}} = I_1\vec{\omega_1} + I_2\vec{\omega_2}$$
NOTE: Here $\vec{L_1} = \vec{L_2}$ as the point of contact lies midway between the centers.
Thus on solving we get,
$$\vec{L_{net}} = 0$$
Hence even if the discs stop rotating (which is not necessary) the angular momentum will still be 0 and hence conserved.
But as pointed out by others, it is not necessary that the final angular velocities be zero as it depends upon their masses and radii. Also the rotational kinetic energy initially would be greater than the final; hence the kinetic energy will not be conserved. 
A: They don’t stop rotating. As friction starts, they will start to rotate around the point of contact. 
A: The two disks are going to exchange an impulse (aa chunk of momentum) that is going to be acting at different radii for each disk resulting in an exchange of different amounts angular momentum. In the end, the disk may not stop spinning, but will spin a compatible fashion (no slip). If you force the disks to stop, then you are violating the conservation
Consider two free floating disks with incompatible rotations, where a single gear tooth at point A is about to go into contact at some point in the future.

The translational and angular momentum of each part is $$ \begin{aligned} p_1 & = 0 & L_1 & = I_1 \omega_1 \\ p_2 &= 0 & L_2 & = I_2 \omega_2 \\ p_{\rm total} & = 0 & L_{\rm total} &= I_1 \omega_1 + I_2 \omega_2 \end{aligned} $$
Total angular momentum is the same regardless of point of measurement since translational momentum is zero. So we might as well measure total angular momentum about the contact point. 
Now the contact happens when points A align. This causes an equal and opposite impulse $J$ to be acting on the two disks.

Regardless of the value of $J$, the result is not only a change in angular velocity $\Delta \omega_1$ and $\Delta \omega_2$, but also the acquiring of translational velocity of the two disk centers $\Delta v_1$ and $\Delta v_2$.
$$\begin{aligned}
  \Delta v_1 & = -\frac{J}{m_1} & \Delta \omega_1 & = -\frac{R_1\,J}{I_1} \\
  \Delta v_2 & = +\frac{J}{m_2} & \Delta \omega_2 & = -\frac{R_2\,J}{I_2}
\end{aligned} \tag{1}$$
Now the translational and angular momentum change of each part is
$$ \begin{aligned}
  \Delta p_1 & = m_1 \Delta v_1 = J & 
  \Delta L_1 & = I_1 \Delta \omega_1 - R_1 (m_1 \Delta v_1) = 0\\
  \Delta p_2 & = -m_2 \Delta v_2 =  -J & 
  \Delta L_2 & = I_2 \Delta \omega_2 + R_2 (m_2 \Delta v_2) = 0
\end{aligned} \tag{2}$$
So the change in total translational and angular momentum is 
$$ \Delta p_1 + \Delta p_2 = J - J = 0 \; \checkmark \tag{3}$$
$$ \Delta L_1 + \Delta L_2 = 0 + 0 \; = 0\checkmark \tag{4}$$
So conservation is valid, regardless of the impulse amount.
Now if the final state requires no slip, or $\Delta v_1 + R_1 (\omega_1 + \Delta \omega_1 ) = \Delta v_2-R_2 ( \omega_2 + \Delta \omega_2)$ then use the step velocities from above to find that
$$ \left( \tfrac{1}{m_1} + \tfrac{R_1^2}{I_1} + \tfrac{1}{m_2} + \tfrac{R_2^2}{I_2} \right) J = R_1 \omega_1 + R_2 \omega_2 \tag{5}$$
which is solved for $J$.
Note that the final motion is found from (1) with $\omega_1^\text{final} = \omega_1 + \Delta \omega_1$ and similarly for all other quantities.
The result is that only under specific conditions either one of the disks can seize rotating after the contact, but not both at the same time. Even when the two disks are identical, in the end, their centers will translate up and down, and they will counter rotate.
The way to make the two disks stop, is if their centers are connected to ground. This can be modeled above with $m_1 \rightarrow \infty$ and $m_2 \rightarrow \infty$. The necessary condition for the disks to stop after impact is then
$$ \frac{L_1}{L_2} = \frac{I_1 \omega_1}{I_2 \omega_2} = \frac{R_1}{R_2} $$
A: Think of two identical  wheels, both spinning clockwise, with the same mass and angular acceleration. If they are pushed together, the front of one will contact the rear of the other. So while they are both rotating clockwise, the front and rear contact points have opposite linear direction. There will be friction at contact until their equal kinetic energy is spent, and they both stop. The axles holding them together provide an external force.
A: After getting confused about the original question, I decided to state it differently, in a way that I hope will be more clear.
You have two planet-size spheres, that each rotate fast, clockwise around galactic north. The total angular momentum of the system is double the angular momentum of one planet.
These spheres are constructed of a special handwaving material that is strong like steel, and conducts heat extremely well. 
Start with the spheres motionless (except spinning) with 1000 meters separating their surfaces. Gravity will bring them together. When they rub against each other, friction creates heat which spreads through them. I postulate that the friction from their mutual spin will create just enough heat to melt both of them, so they coalesce into a single molten blob. What happens to their angular momentum?
And the answer is that it is conserved. Some of it will go to clockwise motion of the liquid. The fastest atoms will tend to be concentrated in a ring at the equator. 
Some of it will go to clockwise rotation of individual atoms. They will rotate clockwise much more than anti-clockwise. 
The total angular momentum will be preserved, because angular momentum is preserved.
A: Angular momentum is conserved but not rotational energy. I agree with @Tom B. that the disks continue to rotate. 
Imagine the two disks rotating with angular velocity $\omega_i$ in vacuum in absence of gravity, so that it is an isolated system. Suddenly they make contact and lock on to each other making one single rigid body. The total system will now rotate about the new center of mass, which is the contact point if the disks are identical. The moment of inertia is increased from $$I_i=\frac{1}{2}mr^2$$ per disk to $$I_f=3mr^2=6I_i ~,$$ where I use the Huygens-Steiner parallel axis theorem. The total angular momentum is $$J=2I_i\omega_i=I_f\omega_f ~.$$ Therefore the angular velocity of the total system is $$\omega_f=\omega_i/3~.$$
The initial total rotational energy is $$E_i=I_i\omega_i^2 ~.$$ The final rotational energy is $$E_f=\frac{1}{2} I_f \omega_f^2= \frac{1}{3}I_i \omega_i^2 ~.$$ The missing energy is added to the international energy of the disk system. It will either be converted into heat, assuming that the disk system's internal degrees of freedom are dissipative.It may of course also lead to a break-up of the system. 
A: What is conserved, is the angular momentum of the whole system. One disk can have an angular momentum which is not conserved -- at any time something from outside can make it spin faster or slower.
We are thinking about a system with just the two disk, with nothing from outside affecting them. They interact and change each other's angular momentum, but the system's angular momentum started out zero and it's still zero afterward.
