Don't be fooled, this problem at first appears trivial, and then impossible.
Before you ask, there is no way to take out any of the resistors from the mesh or to know their values by their markings. The project is commercially sensitive so I'm afraid the discussion is going to have to be restricted to the example given. In the real world problem, the mesh will be very much larger but the resistor values will all be within +/-20% of each other. I've exaggerated the values in the example to make it easier to conceptualise where the currents are flowing.
The problem is simply this. If we know the voltages at all of the nodes, and we know the voltage across the mesh and, crucially, the current flowing from the voltage source, how can the resistor values be calculated?

Here are a few thoughts....
Node or mesh analysis can't provide enough equations for the number of unknown resistors.
If you were to allow the values of the bridging resistors to become infinite, multiple solutions are possible, but this won't be the case in practice.
My guess is that you can't solve this using classical methods because those methods can't accommodate those infinite values and discount them.
Once you add the constraint that no resistor can have an infinite value, I think there is only one solution, but I have no idea how to find it.
Any ideas?

 

Do you require an "exact" solution, meaning that the calculated resistor values give a network where VB1, VB2, VB3 and iV5 are the same as the example network gives to within, say, 6 or 7 digits of accuracy?

You say that all the resistors are the same within 20%. Will you know what that value is when starting the search for a feasible network?

 

You have 8 unknowns (in the figure if all the resistor values are unknown).

You have five mesh equations and three node equations.

 

I was able to find another set of values that satisfy the 4 criteria:

R1 = 208.80858468280957; R2 = 133.72931408370076;
R4 =232.67397559746723; R5 = 789.5680897947801;
R7 = 446.6046060851423; R8 = 240.2470482079432;
R10 = 101.31105474335516; R12 = 336.099432067319;