The point of this exercise is to apply simplifying assumptions and use the rule of thumb to specify the output impedance and input impedance of each part of a circuit so that the driving circuit is not significantly loaded down, i.e. voltage attenuation is less than 10%. In this case you break up the circuit into 2 parts where R1R2 drive R3R4. So you are concerned with Rin and Rout of each. Input of R1R2 is at the top of R1, the output is the node between R1 and R2; the output of R2R3 is the node between R2 and R3.

Now having established what is driving what.... The rule of thumb applied to this problem is Rout for the R1R2 network must be <= 1/10 of Rin for the R3R4 network. The simplifying assumption is that the input impedance that R3R4 sees at its output will be infinite (open circuit); and that the output impedance of the source that drives R1R2 (V) is zero.

What is the input impedance of R3R4 that R1R2 sees? 10K +10K = 20K. So the Rout of R1R2 network must be <= 1/10(20K) =2K. Therefore R needs to be 4K or less because the thevenin equivalent of these 4K || 4K is 2K, which is R1R2's output impedance.