This is an interesting problem, and certainly a model for the forces and torques could be generated. More on this in a moment. However, I think you may run into some issues with the "linear difficulty" curve. The model would tell you the forces, but that doesn't directly correspond to "difficulty." The muscular system can have different efficiencies in converting energy to force dependent upon the orientation of the body. For example, I recall years ago reading about how soccer players throw the ball from the sidelines at something like 60 degrees, much higher than the 45 degrees kinematics would predict as optimal (unfortunately I no longer remember the source) just because the human body is able to throw harder at this higher angle and that extra oomf overcomes the drawback of throwing at a higher angle. Though, perhaps I should add the disclaimer that this is not my area of expertise, so grain of salt and all that.
As a result, while we can model this situation and we can find a progression of hand locations which will result in something like a linear torque progression on the shoulder, it's not clear to me that this will necessarily correspond to a linear progression in the difficulty of the exercise.
With all that said, we can create a simple model to get a quick set of estimates...models can always be made more complicated with additional bells and whistles. So for now, let's assume the person is essentially flattened and consider only 1 arm (so if we use 2 arms, each arm must supply half the forces and torques we find in this model, but simplifying to 1 arm will make the model much simpler).
Roughly speaking, the diagram should look something like this:
I have labeled the stick frame of the person in black, the measurements in yellow, and some forces in green. Clearly this is very simplified, but hopefully will be enough to get started with. Also note that I have labeled the diagram with more information than is strictly necessary to solve a statics problem of this type because it may be useful for your purposes to use somewhat different variables. For example, the angles can be expressed in terms of the height $h$ (which is the height only up to the shoulders by the way), the arm length $L$, and the distance $x_h$ between the hands and feet by using some trigonometric identities...law of sines, cosines and so on.
Now, note that I have not labeled any forces on the shoulder. These are internal to the body and I'll come back to what I think is the right way to get at the torque that the musculature has to apply (realistic models, at least what I found on a quick google search, of the human shoulder can get very complicated very quickly). The problem at hand here is to find the forces supplied by the hands and feet, labeled $F$ and $N$ with appropriate subscripts, such that the body doesn't move.
To do this, I found it convenient to write down Newton's equations for the entire body to impose the condition that the body does not sink through the floor or begin levitating away, and the torque equations to make sure the body doesn't rotate, we are supposed to be remaining still after all. For the torque equations, I chose the feet and the shoulders as my pivot points, though you could have chosen any pair of points you like. The resulting equations are
$$
0=N_h+N_f-mg,\ \ \ 0=F_h-F_f, \ \ \ 0=N_hx_h-mgx_{CM},\\
0=mg(h\cos\phi-x_{CM})-N_fh\cos\phi-F_fh\sin\phi-N_hL\cos(\theta+\phi)+F_hL\sin(\theta+\phi).
$$
These are a little bit messy, but at the end of the day they are only linear equations for the forces we are asking about, and, assuming I've made no mistakes to this point, the solution is given by
$$
N_f=\frac{mg}{x_h}(x_h-x_{CM}),\ \ \ N_h=mg\frac{x_{CM}}{x_h},\\
F_f=F_h=mg\frac{x_{CM}}{x_h}\frac{x_h+L\cos(\theta+\phi)-h\cos\phi}{L\sin(\theta+\phi)-h\sin\phi}.
$$
Again, this is a little bit messy. Perhaps there's another set of variables which would make this look neater, such as eliminating the sines and cosines in favor of the other measures I included in the diagram. Not sure, haven't checked.
Now, we know all the forces acting on the body through the hands and feet and, as a result, the forces the hands and feet are applying to the floor. Assuming the shoulder is supplying all the torque to make this happen and maintain the angle $\theta$ in the diagram, I think it would be reasonable to define the "torque" that the shoulder has to supply as being the sum of the magnitudes of the torques supplied by the hands and by the feet. Admittedly, there could be something wrong with this thinking, and if anyone has another idea, I'd be interested to hear it. If we run with this idea though, we can now write the "torque" supplied by the shoulder (the quotes because it's not really a torque so much as a measure of torques supplied on either side of a joint) as
$$
\tau=mg\frac{x_{CM}}{x_h}\left(|x_h+h(F\sin\phi-\cos\phi)|+L|\cos(\theta+\phi)-F\sin(\theta+\phi)|\right),
$$
where I have used Mathematica to just plug everything in and simplify. I have also left $F\equiv F_f=F_h$ in this expression for reasons described in the edit below.
Finally, let me note again that this $\tau$ is some measure of the effort both shoulders need to exert. If you want a per-shoulder measure, then halving this seems reasonable.
Edit: Let me comment on something I entirely brushed over in the above comments, but is actually important to understanding how this model works. The attentive reader may have noticed that the expression obtained for $F_f$ and $F_h$ is actually ill-defined as both the numerator and denominator of the fraction appearing therein are both zero, though this is hidden by the fact that we are working with too many variables. Essentially, and this can be checked directly, once we have eliminated the normal forces and obtained an equation for, say, $F_f$, the equation takes the form $aF_f=b$ where $a$ and $b$ are zero. As a result, the above equations do not actually determine $F$.
How can we understand this? Aside from this point out answer seems to make sense: in the special case where $x_h=x_{CM}$ we have $N_f=0$ and $N_h=mg$, which is precisely what we would expect when the person is just holding themselves up with their hands alone. One way to imagine this is as follows. For the moment, let's remove from our minds the human body and suppose instead we are dealing with a pair of steel beams that we've welded together at what we have been calling the shoulder. Just placing such a thing on the floor, I think it might be intuitive to expect that there would be no need for horizontal forces acting on the ends of the beams.
At the same time, however, we could imagine ourselves getting on the floor to do, say a pushup or a plank. While doing so we always have the option to apply pressure to our hands and feet which is parallel to the floor. Doing so would lead to greater-than-necessary exertion, but we could do so without actually moving in any way. I think this is precisely what's happening here: the math is telling us that any amount of horizontal forces we would like to apply would work out fine, but we actually don't need any of them, at least in this simple model.
This is why I went back and edited the expression for $\tau$ above to just have $F$s in it. We can plug in any value we like for $F$: we are always free to exert more effort for no reason. There will, however, be a minima to this function which will correspond to the minimal amount of effort. And by the way, $F=0$ is not necessarily the point of minimal effort because we are dealing with a function whose basic form is $|aF+b|+|cF+d|$ where the constants $a,b,c$, and $d$ could be either positive or negative (we know it has a minima because $F\rightarrow\pm\infty$ causes the function to reach positive infinity on both ends, implying a minima somewhere in the middle by continuity). This is, I think, the mathematical realization of the observation that if we were to do a pushup or something, most people would agree that they do, in fact, apply some non-zero horizontal force...the minima of effort actually corresponds to applying some non-zero horizontal force in general.
