Bingo, found my mistake. I was thinking in DC so I simply added resistance with impedance. I forgot that it has to be vector sum (sqrt R^2 + X^2).Based upon the above data, I calculated a value of ≈4.8μF for a series capacitor to give the same current at 245V as you were getting with the dimmer at 205V.

I calculated an equivalent resistance of 1.001Ω for the motor at 205V, giving a current of 205mA.

The impedance to give the same current at 245V is then 1.196kΩ.

The vector impedance of the series resistor and capacitor is the square-root of the sum of the squares of their impedances.

Solving that gives a desired capacitance impedance of 654.9Ω.

At 50Hz this is provided by a 4.8μF cap.

My LTspice simulations verified this value.

But note that the accuracy of the meter is suspect for the chopped sine-wave that the dimmer generates to reduce the voltage.

Don't know if it measures true-RMS or not for such a waveform.

I've cobbled together a circuit with two paralleled lightbulbs in series with the fan and by changing the wattages of the bulbs was able to get it to run well (the cold resistance of the paralleled bulbs was 120 ohms, when fan switched on the bulbs glowed dimly and I measured 55V across the bulbs at just under 200mA so hot resistance of about 1kohm). This made it run a bit slower than ideal but it's a reassuring proof of concept until I can get the capacitive circuit set up.

I've determined that I will have to buy capacitors on eBay as local availability is spotty. Can you suggest capacitors to buy? If the correct value is about 4.7uF I was thinking of buying a pack of 10x 1uF 400V caps. Then I could parallel them until I get a comfortable speed out of it (giving me a range of 1-10uF). What do you think of that strategy? Am I correct in assuming that more capacitance = less impedance = more speed?