How to choose a resistor value for Wheatstone bridge such that it is a balanced circuit?
As commented you are probably making confusion between Kelvin connections and Wheatstone bridges. They are completely different things!
A Wheatstone bridge is simply a glorified two-voltage-divider system, really.



simulate this circuit &ndash; Schematic created using CircuitLab
One divider on the left, one divider on the right and a null detector in the middle (in the original construction). When it was invented you put your X resistor at, like R1 position and tuned R2 or R4 until you got 0V between branches. R3 would set the range, more or less. The advantage is that R1 and R3 can be of different orders than R2 and R4 (it's not practical to make a, say, 10 ohm variable resistor).
These days the Wheatstone is mostly used as an amplifying device for transducers with multiple resistor varying (i.e. strain gauges and similar).
A Kelvin connection (or Kelvin sense or 4-wire resistor measurement) is a way to avoid errors due to connecting lead resistances:



simulate this circuit
The idea is that the 'wire joints' are made directly on the resistor (current shunts have actually 4 terminals precisely connected in that way).
With a voltammetric measurement you force a known current in the resistor and measure the drop (Ohm's law follows). However the measurement leads themselves have some resistance (on the order of 0.1 ohm, typically) that would get included in the measurement with a standard ohmmeter. Using two different leads and and high impedance voltmeter you can avoid the error since on the sense leads circulates no current.
Now, measuring a resistor with a Wheatstone bridge and a single ended ADC like the one on the Arduino simply can't be done. You need a differential amplifier for that. And it's actually trickier than it seems since the error and tolerances propagate really fast (also, by the way, the Wheatstone bridge is not linear with only one varying resistor).
1 to 40 ohm with 0.1 ohm precision is quite feasible and not too hard with a Kelvin sense. The main issue is the power you dissipate on the resistor, that could be a problem. Assuming you are not going to use a dedicated amplifier (they make them), if you wanted 2.5V with a 40 ohm resistor you would need 62.5mA and dissipate about 156mW on the resistor in the process.
For the resolution: assuming 2.5Vref and a 10 bit ADC (not sure if the Arduino one is 10 or 12 bits), one LSB is about 2.44mV. 0.1 ohm on a 62.5mA current gives 6.25mV, so it's about 2.5 counts of converter. Could be better but on paper you have the resolution needed, oversampling a lot would help if possible.
As for the precision… well, you didn't say. Precisely is not exactly… precise. In this application the main sources of error are the constant current generator and the ADC Vref (the Vref generator internal to the MCUs usually is quite poor).