Capacitor value substitution

First you have to know the leakage current limit for your class of equipment. For IEC60950 IT equipment it is 250uA. Then the leakage current will be given by: Cy = (Il/ ωV) – Ciw where Cy is the Y-capacitor value, Ciw is the interwinding capacitance of your transformer (specified or measured) ω is the line frequency and V is the line voltage.

Of course you need to build in margin for all the tolerances involved (check temperature variation as well) and you need to measure the leakage current to be sure the calculation is correct.

Ignoring implementation specifics downstream and looking at balancing only:

The current through the resistors should be at least 10x the capacitor leakage current to provide a stiff division of the input voltage (i.e. the capacitor leakage current is insignificant relative to the resistor current).

Calculate this at \$Vin\$ (min), so 400v/1.67mA yields close to 240k\$\Omega\$, so two 120k\$\Omega\$ resistors will do the trick.

Power in each device is calculated at \$V_in\$ (max), so from \$V^2/R\$ I get 4.5W for each resistor.

I would probably use 10W devices.

Note that the resistance is high enough that it should not affect normal downstream performance (but that is application specific which I am not trying to answer)