Step by step:
The current, from left to right, through \$R\$ is
$$I_R = \frac{V_{SRC} - V_{O2}}{R} $$
The current, from left to right, through the left-most 10k resistor is
$$I_{10k} = \frac{V_{SRC}}{10k\Omega} $$
KCL at the input node yields
$$I_S = I_R + I_{10k}$$
Using the well-known inverting op-amp gain formula, the two op-amp cascade has a gain of
$$\frac{V_{O2}}{V_{SRC}} = (-\frac{40k}{10k}) \cdot (-\frac{20k}{10k}) = 8 $$
Now, set \$I_S = 0\$ and solve.
A rewarding exercise is to solve for the input resistance seen by the input voltage source:
$$R_{IN} = \frac{V_{SRC}}{I_S} = \frac{V_{SRC}}{I_R + I_{10k}} = \frac{1}{\frac{1}{10k\Omega} - \frac{7}{R}}$$
Note that the input resistance is positive for \$R > 70k\Omega\$, is negative for \$R < 70k\Omega\$ (the circuit supplies power to the voltage source), and is 'infinite' (open circuit) for \$R = 70k\Omega\$
There's actually a really straightforward way to approach this, if you don't mind using a matrix calculation (Matlab, for example).
In the traditional circuit analysis, the unknown variable is a vector of voltages corresponding to the ports. Choose one of the ports as a reference, I'll use the bottom one since that's where we usually put ground on schematics)
$$\mathbf{x} = \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \right] = \left[ \begin{array}{c} V_{15} \\ V_{25} \\ V_{35} \\ V_{45} \end{array} \right]$$
For each port you either have a trivial voltage equation setting that port voltage to the applied input, or a conservation of current equation (net current out of a disconnected port is zero).
For excitation connected to ports 1 and 2, the trivial equation is
$$x_1 - x_2 = V_{input}$$
and the KCL equations are (using conductance G which is the reciprocal of resistance R)
(port 3) $$G_2 (x_3-x_1) + G_5 (x_3-x_2) + G_8 (x_3-x_4) + G_9(x_3) = 0 $$
(port 4) $$G_3 (x_4-x_1) + G_6 (x_4-x_2) + G_8 (x_4-x_3) + G_{10}(x_4) = 0 $$
(port 5) $$G_4 (-x_1) + G_7 (-x_2) + G_9 (-x_3) + G_{10}(-x_4) = 0 $$
Because there is one equation for each port and one unknown for each port, you immediately have a system of linear equations. As long as this system isn't degenerate (which it never is with the fully-connected graph of resistors, if all resistor values are finite) there will be a unique solution.
$$\left[ \begin{array}{c c c c} 1 & -1 & -0 & 0 \\ -G_2 & -G_5 & G_2+G_5+G_8+G_9 & -G_8 \\ -G_3 & -G_6 & -G_8 & G_3+G_6+G_8+G_{10} \\ -G_4 & -G_7 & -G_9 & -G_{10} \end{array} \right] \mathbf{x} = \left[ \begin{array}{c} V_{input} \\ 0 \\ 0 \\ 0 \end{array} \right]$$
A simple matrix inversion, which is easily automated, yields all the port voltages in terms of the resistances.
Unlike the traditional analysis, you want the resistances, so we'll organize the matrix multiply the other way, and also include the excitation current:
$$G_1 (x_1-x_2) + G_2 (x_1-x_3) + G_3 (x_1 - x_4) + G_4 (x_1) = i_{exc}$$
$$\left[ \begin{array}{c c c c c c c c c c} x_1-x_2 & x_1-x_3 & x_1-x_4 & x_1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & x_3-x_1 & 0 & 0 & x_3-x_1 & 0 & 0 & x_3-x_4 & x_3 & 0 \\ 0 & 0 & x_4-x_1 & 0 & 0 & x_4-x_2 & 0 & x_4-x_3 & 0 & x_4 \\ 0 & 0 & 0 & -x_1 & 0 & 0 & -x_2 & 0 & -x_3 & -x_4 \end{array} \right] \left[ \begin{array}{c} G_1 \\ G_2 \\ G_3 \\ G_4 \\ G_5 \\ G_6 \\ G_7 \\ G_8 \\ G_9 \\ G_{10} \end{array} \right] = \left[ \begin{array}{c} i_{exc} \\ 0 \\ 0 \\ 0 \end{array} \right]$$
Now you'll have to take all of those results together and get your conductances, which lead easily to resistances.
Let's calculate the size of that final step. There are (k choose 2) total resistances, and you have (k choose 2) ways to energize the system, with (k-2) intermediate voltage measurements from each. Plus one measurement of energizing current for each, if you choose to use it. That's (k-1)*(k choose 2) equations, which should be plenty for solving (k choose 2) unknown resistances. The system will be overconstrained, but you'll have some measurement error, so a pseudo-inverse will give the set of resistances that is most consistent (in a least-squares error sense) with your measurements.
You can presumably get away with just using the equation for input current for each of the (k choose 2) excitation patterns, and have enough equations, but without redundancy any measurement inaccuracy can lead to a big discrepancy in the end... the redundant equations help protect against that.
The key to a systematic approach is to not write loop equations at all, let the matrix algebra system derive them automatically.
Best Answer
I'll show you the general mathematical approach and you'll proceed.
For the purpose of the tutorial lets take the upper part of your circuit. I rearranged it a bit and you need to check that there is no mistake there:
simulate this circuit – Schematic created using CircuitLab
Lets define the current through each resistor as flowing downwards. In this case, the following matrix is the complete representation of the circuit (and if it isn't - find additional constraints and it will be):
$$\begin{bmatrix} 1&-1 &-1 &0 &0 &0 &0 &0 &0 \\ 0&1 &0 &-1 &-1 &0 &0 &0 &0 \\ 0&0 &0 &1 &0 &-1 &0 &-1 &0 \\ 0&0 &0 &0 &1 &0 &1 &0 &-1 \\ 0&0 &1 &0 &0 &1 &-1 &0 &0 \\ & & & & & & & & \\ 0&R_2 &-R_3 &R_4 &0 &R_6 &0 &0 &0 \\ 0&-R_2 &R_3 &0 &-R_5 &0 &R_7 &0 &0 \\ \end{bmatrix}* \begin{bmatrix} I_1\\ I_2\\ I_3\\ I_4\\ I_5\\ I_6\\ I_7\\ I_8\\ I_9\\ \end{bmatrix} = 0$$
Now just use any program which is capable of solving the systems of linear equations and you can get all the currents in terms of any pair of currents (lets say \$I_1\$ and \$I_2\$ for the sake of this tutorial). You can even solve this system by hand (are you crazy???).
Afterwards, you'll have to do few more calculations because you did not provide any boundary conditions. The additional equations are:
$$V_{in}-V_{out1}=I_1R_1+I_2R_2+I_4R_4+I_8R_8$$
$$and$$
$$V_{in}-V_{out2}=I_1R_1+I_3R_3+I_7R_7+I_9R_9$$
EXPLANATION:
The first five equations in the matrix are KCLs for inner nodes (sum of currents in each node is zero), and the last two equations in the matrix are KVLs for inner loops which introduce voltage constraints (sum of voltages on each closed loop is zero).
Once all the currents are known in terms of \$I_1\$ and \$I_2\$, and the boundary conditions are given, the last two equations (not in matrix) may be solved to obtain \$I_1\$ and \$I_2\$.
Once you know all the currents (and consequently all the voltages) in the network - it is solved.
IN CASE YOU STILL WANT SIMPLIFIED SCHEMATICS:
You won't be able to separate the branches leading to
out1andout2. The reason is that they share a common nodes, and the change in boundary conditions on one node will alter the currents in the branch of the other too. However, you might find some parts of your network which can be simplified. For example, in your original schematic, it seems like you might be able to represent \$I_{A}\$ and \$I_{B}\$ in terms of \$I_8\$ and \$I_9\$ in a quite simple manner, and reduce the bottom ladder of the resistors to just a few resistors. However, you still need to solve the above equations as prerequisite.NOTE:
If either
AorBcan be defined a constant, it can simplify the issue a lot.