Following on from Peter's answer I will add some more suggestions although I also recognise some ambiguity (or underdetermination) in the problem - some will be identified at the end of the discussion.
There seems to be two aspects to the physics of this:
- Determine the forces at $[x_i,y_i]$ for all the N points on the box
- Determine whether there will be some non-zero moment to the box.
The equations of statics that are basic to this are:
$\Sigma F_x = 0$ - the x axis forces sum to zero
$\Sigma F_y = 0$ - the y axis forces sum to zero
$\Sigma F_z = 0$ - the z axis forces sum to zero
$\Sigma M_a = 0$ - total moment about any axis is zero for equilibrium.
Other forces include the reaction $V_i$ forces at each $[x_i,y_i]$.
One problem is that this scenario is statically indeterminate - meaning that the distribution of forces using only statics will not give a complete answer. Some other form of approximation or data is also required. The method of sections (briefly in Wikipedia here) and perhaps the moment distribution method are worth studying. These methods are for solving general engineering problems of this type (like the truck itself maybe). I do not know whether there is freedom to simplify some of these properties in your model, thus resulting in a much simpler "force distribution model" inside the box.
One suggestion might be to consider a "truss approximation" to the floor of the box (even the entire box) with the truss joints corresponding to the reaction points of the question.
From the perspective of the non-zero moment of the box it is not clear where the freedom of movement is coming from. For example if each of the N points were considered a high stilt then the configuration might not be stable, for example.
Looking at the simplest possible example of this case we have a(n essentially massless) beam of length L sitting on two (high) structs at A and B with each point interior to the beam, maybe A is distance a, B distance b from the LHS. Let a force $F_1$ be acting on the LHS downwards, and let $F_2$ be acting on the RHS downwards (perhaps from edges etc). Then the
LHS moment = $aF_1$
RHS moment = $(L-a)F_2$
For stability we require RHS>LHS ie $(L-a)F_2 > aF_1$ ie $F_2 > (a/(L-a))F_1$.
Clearly these equations can and must be generalised further depending on the exact level of assumptions one includes in the full model (massive box, height/strength of structs, interior mass distribution in box, strength of box walls, side forces, friction, ...).