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)