Electronic – What’s the purpose of the ‘positive feedback resistor’ in the circuit

Something like this should do you. Note that the input common mode voltage range of the LM358 amplifier only goes to 3V so you can't just buffer a divider with voltage follower (which would be the easiest way, and would typically work around room temperature, but this is engineering so we have to consider worst-case and temperature).

The output range, with load of more than 2K, goes to within 3.5V so it's fine.

The input divider has a ratio of 0.50 (and it loads the input with 40K)

The amplifier has a gain of 1 + 10K/27K

Total gain is 0.5 (1+10/27) = 0.685, so 4.8V -> 3.28V.

^{simulate this circuit – Schematic created using CircuitLab}

Ideally, pick values such that R3||R4 ~= R1||R2 to cancel the effect of input bias current. In the above schematic they are about 30% different.

Both inputs of the DUT are grounded. There shouldn't be any current flowing through the 10ohm resistor and the 100Kohm resistor, as there is no voltage source.

Wrong - the internal voltage noise source of the inputs is always present and that is what is being tested by putting the amplifier into a gain of 10,000.

Here is a more general view of op-amp noises: -

The noise source (implied by the circuit in the question) that is specifically being tested for is labelled \$e_n\$ in the diagram above.

These number do not exactly match with the description of the circuit (0.1Hz - 10 Hz), but perhaps it's close enough.

They are using two filters at 16.8 Hz which together produce a 3dB point of 10.8 Hz. Having said that this will have a noise bandwidth of 12 to 13Hz (not an exact science) so it's a little too much beyond the 10 Hz limit in my book - this calls into question the guy who designed it.