Confused by very straight-forward "garage-door" problem I'm somewhat embarrassed to ask this here as the problem seems to be so trivial and we (yes, I have been asking many people around me to pitch in - to no avail) always reach the same conclusion, which however is totally contradicted by reality, so ...
(Spoiler ... I think I've found the difference between reality and the problem description below.  I'll add it to the bottom of the question though - no use hiding my mistakes ;-) )
I have a garage door that stays inside of the walls.  I have no idea whether this type of door has a particular name in English, so to ensure we have the same understanding I'll add a picture and some explanation:

The garage door is one solid panel (typically of wood or with an internal metal construction) and has at the left and right sides a wheel at the top running inside a fixed (typically via some construction to the ceiling) horizontal rail and at the bottom a wheel running inside a fixed (to the side-wall) vertical rail.
Because of the rigidness of the door, I believe we can reduce this problem to 2D, so basically to the cross-section view given by the attached image.  As the wheels have some slack in the rails, they can roll inside them and hence I also think we can assume they won't provide any force in the direction of the rail, only perpendicular to it.
In most installations (as in mine) there will be a counter-weight installed which using cables pulls straight up at the point where the lower wheels are connected to the door.  Obviously the counter-weight would be such that the (manual) amount of force needed to open or close the door would be minimal ...
And this is where theory and reality clash.  If I call the mass of the door m and that of the counter-weight M, all my attempts indicate that choosing M = m/2 would have this door stand still in any position/angle that it can take, i.e. that it would never further lower or further rise out of itself if put (and shortly held) in that position.
Reality however shows that there is a point where the door is in balance and doesn't move on its own (if fully closed is 0 degrees, then my door seems in balance around 30 degrees - but I guess that might depend on the counter-weight), but that below this point (so more closed), the door further closes automatically and above that point (so more open) the door further opens automatically (and my arms can tell you that the force needed to counter this grows significantly the further you go from that balance point).  That seems to indicate that the force needed to counter the door's own weight should (unlike the constant weight of the counter-weight) be a function of the angle the door is at.  But as said, this is not what all my attempts to calculate that force result in.
Hence the question: can someone give a formula for the force one would have to supply vertically at the bottom attachment point of such a garage door in function of the weight of the door and the angle it is positioned in?  And also explain how you get to that formula.  Or otherwise explain why reality and simple mechanics calculations seem to clash here?
Adding some more info as it seems some people think the counter-weight would be something very special:

As mentioned a steel cable starts at the bottom wheel at each side of the door, runs over a pulley at the top, so that the cable pulls vertically on that bottom wheel and then via a second pulley to the counter-weight.  In my case there is indeed a pulley for each cable attached to the weight and the cable then runs up and is fixed to the ceiling.  This just halves the distance the weight needs to travel and the force it applies to the door with its weight, but still that force will be constant over the travel of the weight (and hence the door), right?
Thanks in advance for help saving my sanity.
PS: the border case with the door fully closed also gives a hint that somehow the angle must play a role as it's easy to realize that in this position you will never get the door to open unless you also give it a sideways push.  So somehow that angle must have made the F needed to open the door infinite (fully countered by V), right?  Of course, that is not the F needed to keep the door in equilibrium, but still it raises that feeling that the angle does play a significant role, no?
==== NORMALITY HAS BEEN RESTORED ====
Or how oversimplification probably caused all my problems above ...
Looking a bit closer at the door, I started to realize that the center of mass of it is actually not in the middle between the wheels.  Hence a more correct sketch of the real situation would be:

And when I then do a static analysis, I do get a $tan(\alpha)$ in the result and with the counter-weight slightly heavier than half the weight of the door, I also get a point between the extremes where the door would balance.
Conclusion: my original problem description is likely oversimplified in that it assumed the center of mass to be on the axis between the wheels.  The further that center of mass moves away from this axis, the more the position of the door plays a role.
 A: There are many different ways for this to work, so without actually looking at your garage door there is no way to give the correct answer.
I'll write $L$ the length of the door (length of the purple rod). I'll assume for simplicity that its center of mass in in the middle of the door. I'll also note $x$ the vertical displacement of wall joint, in particular $0\leq x\leq L$ with $L$ the length of the door.
The principle of virtual work says that:
$$
Fdx = dE
$$
with $F$ the force needed to apply at the moving joint to move it by an increment $dx$ and $E$ the energy of the system, in this case the potential energy of the door and of the counterweight, of mass $m,M$ respectively. You thus need only calculate $E$ as a function of $x$ and take the derivative to get $F$, which is what you want.
From geometry, you get:
$$
E = Mgx-mg\frac{x}{2}
$$
which gives your result.
Your observation indicate the presence of a single unstable equilibrium $x_0$. In terms of energy, this translates as a single global maximum. The simplest way I could think of is to modify the potential energy of the counter weight by making it piecewise linear rather than completely linear. Physically, the counter weight would be allowed to roll on a incline of variable slope, or something analogous. You could imagine a change of slope in the potential energy.
$$
E = \begin{cases}
-\frac{mg}{2}x+s_1Mgx & x<x_0 \\
-\frac{mg}{2}x+s_2Mgx & x>x_0
\end{cases}
$$
To reproduce your observation, you just need $s_1M< \frac{m}{2}<s_2M$. T'm pretty sure that your garage door must be more complicated than that. In general it is hard to guess what is going on, without knowing about the full apparatus, especially for the counterweight.
Hope this helps.
A: 
From the work that done by the force $~F+M\,g~$ and by the force $~m\,g~$ you obtain
$$\int(F+M\,g)\,y\,dy=
 \int  (-m\,g)\,s\,ds\tag 1$$
with $~s=\frac L2\,\sin(\psi)\quad y=(a-L\sin(\psi))\Rightarrow$
$$ \left( F+Mg \right)  \left( a\psi+L\cos \left( \psi \right) 
 \right) =\frac 12\,mgL\cos \left( \psi \right) 
$$
solve this equation for force $~F~$   you obtain
$$F=F(\psi)
$$

with
$$[a=2,b=1,m=80,,M=32,g= 9.81]~,L=2.23~[m]$$

A: 
And this is where theory and reality clash. If I call the mass of the door m and that of the counter-weight M, all my attempts indicate that choosing M = m/2 would have this door stand still in any position/angle that it can take, i.e. that it would never further lower or further rise out of itself if put (and shortly held) in that position.

Can we see your analysis? Doing a basic static moment analysis at the upper wheel, I get
$$
F_{H}\tan(\theta) = \frac{1}{2}M_{d}g - m_{cw}g
$$
where $F_{H}$ is the horizontal reaction force, $M_{d}$ is the mass of the garage door, and $m_{cw}$ is the mass of the counterweight. The angle is measured from the horizontal, so that $\theta = 0$ when the door is fully open and we see the sensible result that no horizontal reaction is needed in that case. The formula above was found by solving for the reaction force (you call it $H$ in your picture) that would be necessary to give zero moment at that upper wheel for an arbitrary mass counterweight (not assuming half the door mass). As noted below, in general the door is not in equilibrium, so the real reaction force would be different and require a dynamic analysis.
Clearly for the special case of the counterweight equaling half the mass of the door, the horizontal reaction is unnecessary and the system balances for any angle. Otherwise, such angular balance would require a horizontal reaction, which would then push the door (there is no other horizontal force available) and thereby change the angle, meaning the system is not in static equilibrium.
Given this, I am wondering if the counterweight is a bit less than half the mass and this means that the weight of the pully system is non negligible. Beyond a certain point in one direction, the total mass is more than half the door mass, while beyond that point in the other direction, the total mass is less than half the mass. This variation in mass around the singularity at half the door mass would lead to the behavior you are seeing. This would show up in the above equation as $m_{cw}$ being a function of the angle $\theta$
A: I think the key here is that as the door opens, some of the weight of the door is held up by the top track and not by the counterweight. When the door is fully closed, the total force pulling up at the front edge must be equal to the full weight of the door to lift it. But when the door is fully open, only half the weight of the door is at the front edge, while the other half is at the back edge. A counterweight weighing more than half the weight of the door will hold the door open, but won't be enough to open it when closed.
The key empirical detail is how exactly the top wheel of the door starts to take the weight of the door. As the top wheel moves from its initial position, it bears some of the weight of the door, and the remaining force needed to keep lifting the door depends on that transition. In the closed configuration, both the top and bottom wheels on the door may bear weight anywhere between 0 and m, but both bear a weight of m/2 when the door is open by any amount - how that transitions occurs is unclear, but I believe  it's a key physical detail of this system.
A: The good news is that I think I found the source of where the dependence of the support force (from the counterweight) must vary by angle to get balance even for the simplified case initially presented.
The answer is friction of course.
Even though the door is on rollers, there is enough grease and grime usually on these rollers to present some friction, significantly affecting the support force required at shallow angles (measured from vertical for me).

In the image above, $A_x$ and $B_y$ are the rail reaction forces, and $A_y = \mu A_x$ and $B_x = \mu B_y$ the corresponding sliding friction forces. The counterweight force is $S$ and the weight is $W$.
The force/moment balance equations are
$$\begin{aligned}
 A_x + \mu B_y & = 0 \\
 S + \mu A_x + B_y -W &= 0 \\
 \tfrac{L}{2} \sin \theta ( B_y -S-\mu A_x) + \tfrac{L}{2} \cos \theta ( A_x - \mu B_y) & = 0
\end{aligned}$$
The solution for the support force is
$$ \boxed{ S = \frac{W}{2} \left( \frac{2\mu \cos \theta - (1+\mu^2) \sin \theta}{ \mu \cos \theta - \sin \theta} \right) } $$
The solution looks as follows for $\mu=0.1$

The above shows the opposite effect, where when the door is almost open at $\theta = 11°$ about, no support force is needed as friction is sufficient to hold the door up. But with the constant presence of the counterweight force, this means the door will try to open on its own, which is not what happens in real life.
The analysis is a bit complex because as soon as the door tries to open, the direction of friction switches. This results in a different curve that grows higher at lower door angles.
I have added below two sets of friction values, $\mu=0.02$ and $\mu=0.1$, as well as the two friction directions for a total of 4 curves.

Now this means to actually open or close the door completely you need to overcome friction whose effects grow larger the closer the door is to close.
This matches expectations as it is both hard to close and hard to open the garage door due to friction. But midway things can be balanced out and the counterweight is sufficient to support the door, and friction is less important as the curve approaches the $S/W = 0.5$ flat midline which is the ideal solution w/out friction.
