Well, I've just done a bunch of maths, and come to the conclusion that a syphon system won't work.
Code:
V_outer = volume of water in the outer container
V_outer,0 = initial volume of water in the outer container
V_inner = volume of water in the inner container
V_inner,0 = 0 (no water in there initially)

A_outer,internal = area of the outer container, disregarding space occupied by the inner container
A_inner,external = area of the inner container that it occupied in the outer container 
A_inner,internal = area of the inner container that can hold liquid (ie external area minus wall thickness)

d = density of water

M_inner,initial = mass of the inner container at the start (resin + container + weights)
M_inner = mass of the inner container with whatever water it has in it. 
        = M_inner,initial + V_inner*d

h_outer = height of the water in the outer container, relative to the bottom of the outer container
h_inner = height of the water in the inner container, relative to the bottom of the inner container
        = V_inner/A_inner,internal 
h_diff  = difference in height between the water in the outer container and the inner container
h_print = height of the printing surface (ie top of the water in the inner container) with respect to the bottom of the outer container
h_sink  = height of the bottom of the inner container with respect to the water in the outer container - how far the inner one sinks.
Okay, so now we need some equations to relate these. Obviously what we want is for h_print to stay constant. We can freely adjust the container areas, but not really anything else.

First up, how far down does the inner container sink in the outer one? Well, we know its mass, and we know the density of the water, and we know its external area. Therefore we can say:
Code:
h_sink = M_inner / (d*A_inner,external)
       = (M_inner,initial + V_inner * d) / (d*A_inner,external)
We know the height of the water in the inner container, so now we can find h_diff.
Code:
h_diff = h_sink - h_inner
       = (M_inner,initial + V_inner * d) / (d*A_inner,external) - V_inner/A_inner,internal
If we know the height difference and we know the height of water in the outer container, then that's all good. We can easily compute h_print. So, what's the height of water in the outer container? It's not just volume divided by area because that doesn't take into account volume displaced by the inner container. However, if we add that volume then it'll all work.

Code:
h_outer = (V_outer + (h_sink * A_inner,external)) / A_outer
Now, putting that all together and using [V_inner = V_outer,0 - V_outer] we can solve:

Code:
h_print = h_outer - h_diff
        = (V_outer + (h_sink * A_inner,external)) / A_outer - (M_inner,initial + V_inner * d) / (d*A_inner,external) + V_inner/A_inner,internal
        = (V_outer,0 + M_inner,initial/d) / A_outer - (M_inner,initial/d + V_outer,0 - V_outer) / A_inner,external + (V_outer,0 - V_outer)/A_inner,internal
        = V_outer,0/A_outer + M_inner,initial/(d*A_outer) - M_inner,initial/(d*A_inner,external) - V_outer,0 / A_inner,external + V_outer/A_inner,external + V_outer,0/A_inner,internal + V_outer/A_inner,internal
So, what's a constant here and can be ignored? Well, V_outer,0 is clearly constant. All the areas are constant. M_inner,initial is a constant. All that's left over is V_outer! So, rearranging with that we have:

Code:
h_print = V_outer * (1/A_inner,internal + 1/A_inner,external) + <a whole bunch of constants>
The problem here is that for constant print height, we need to make that term constant. We can't keep V_outer constant because the whole point of this design is that you funnel water from the outer container to the inner one. Therefore the only option is to multiply it by zero - which basically means that A_inner,internal and A_inner,external are going up to infinity. Clearly this is not a very practical concept, and even if it was it wouldn't actually work (it keeps the print height constant, but the idea is that the inner container falls as it fills with water. If it's infinitely large it's never going to move much at all).

Back to Mike's idea, which I think might work.