I have a suggestion. In fact, I've had this idea on the backburner for some time.
Question: Given a triple of permutations $\theta=(\alpha,\beta,\gamma)$, with $\alpha,\beta,\gamma \in S_n$, does there exist a Latin square that admits $\theta$ as an autotopism?
(If you're an algebraist, take the same question and replace "Latin square" with "quasigroup".)
I nearly went bonkers answering this question up to $n=17$ for this paper.
While algorithmic methods would take big chunks out of this problem, there would always be some cases that wouldn't work. Backtracking algorithms would sometimes paint themselves into a corner early on, and take virtually forever to escape. And, even if they did work, as soon as I resolve all cases for some value of $n$, it left open the $n+1$ case.
Why this is suitable for crowd computing:
Answering an instance of this question is much like solving a Sudoku problem. All the user has to do is input numbers in a matrix and the computer can check that there's no clashes.
Humans have an advantage over computers: they will be able to see that they painted themselves into a corner early on.
An individual question is not that hard (but there's a lot of them).
Once you have a solution, it's straightforward to check that it's correct, and can act forever as a "certificate" for a given $\theta$.
I foresee implementing this as a puzzle, where the user is presented with a $n \times n$ matrix, with some boundaries (representing the cycles of $\alpha$ and $\beta$) and they place in a symbol from $\{1,2,\ldots,n\}$ into any empty cell. Given that entry, the computer generates the orbit under the action of $\langle \theta \rangle$, thereby filling in some more cells. From the user's point of view, it looks like the numbers "wrap around" and orbits also "pass through" walls in the matrix.
If I'm not mistaken (but I often am), the physicists already have a rather simple way of defining the center of mass. But I don't think you can do it with just sets. You have to associate a mass with each set. The critical axiom is simply the one we all know:
If $A$ and $B$ are disjoint sets with masses $m(A)$ and $m(B)$ and center of masses $c(A)$ and $c(B)$ respectively, then the mass of $C$ is $m(C) = m(A) + m(B)$ and the center of mass of the set $C = A \cup B$ is given by
$$
c(C) = \frac{m(A)}{m(C)}c(A) + \frac{m(B)}{m(C)}c(B).
$$
You do need one more axiom to get started somehow. I believe physicists like to start with point masses (where the definition of the center of mass is easy) and then view a body as a limit of point masses. That's more or less what Liviu has proposed. But it also suffices to say that the center of mass of a square or cube is its geometric center. Or, more generally, the center of mass of any set with sufficient symmetry is its center.
Of course, if you really want arbitrary shapes, then you do need a countable version of the first axiom. But I think that's all you need. Note that this approach allows for bodies with different and even non-constant mass densities.
ADDED (in response to fedja's edit): It's worth noting explicitly that my answer above requires no notion of volume or choice of measure (such as Lebesgue measure) on the ambient space. It works on any length space.
But I don't see any way reduce this to just geometry (and not physics) without a notion of volume. In essence, you do it by just assuming all objects have the same constant mass density, so the mass is essentially equal to volume.
More generally, there has to be a way to measure the relative size of two sets, in order to determine the center of mass of the union of the two sets. If I understand correctly, axiom (4) in the question is an attempt to set this up.
[PREVIOUS DISCUSSION REPLACED BY THE FOLLOWING]
But for me it seems simpler to define a notion of size first and define the center of mass. And, as alvarezpaiva points out in a comment to Liviu's answer, valuations provide the appropriate setting, especially if we restrict to convex polytopes, which are objects that I believe the Greeks understood pretty well. This also allows us to avoid any issues of having to work with infinite sums or unions.
Here, a valuation $f$ is a finitely additive function on the space of convex polytopes. In other words, given polytopes $A$ and $B$,
$$f(A \cup B) + f(A \cap B) = f(A) + f(B).$$
The first observation is that the "critical axiom" stated above is equivalent to saying that $C \mapsto m(C)c(C)$ is a valuation. However, Monika Ludwig showed in her paper Moment vectors of polytopes that the only $R^n$-valued measurable valuation on convex polytopes that behaves appropriately under affine transformations is the volume of the polytope times the standard center of mass.
Ludwig also showed in her Advances article Valuations on polytopes containing the origin in their interiors that any real-valued measurable $SL(n)$-invariant valuation homogeneous of positive degree must be a constant times volume. So it is reasonable to assume $m$ is volume. This therefore implies that $c$ must be the standard center of mass.
Moreover, if you examine Ludwig's proofs, you will see that although they are quite nontrivial, the technology used was arguably within the grasp of the Greeks.
Best Answer
A standard problem of this type is, can one draw uncountably many non-intersecting, non-degenerate figure-eights in the plane? The problem is trivially "yes" for circles, rather than figure-eights, so I found this problem surprising when I first saw it.