Okay, this one requires some application of Ohms law. Adding a resistance in series will not increase current flow it will decrease. To increase current flow the resistance must be put in parallel, and will have to be a large power resistor.
Each headlight drawing 55 Watts at 12 Volts is about 4.5 Amps. This means that the effective resistance of this bulb at 12 volts is 2.66 ohms. Therefore 2.66 Ohms is what the bulb out circuit wishes to see.
The HID resistance of the 35 Watt load is 4.1 Ohms. To make this look like 2.66 ohms to the bulb out circuit, we will need a 7.5 Ohm resistor in parallel with the 35 Watt load to present a 2.66 Ohm load to the bulb out circuit. This will need to be a 20 Watt resistor.
Seeing as how I have been on the software side of the world for about 15 years, I no longer have catalogs to see what the standard available values are. You will need to look it up and get as close as possible.
My calculations were based on 12 volts, when in reality the system will apply about 13.5 volts most of the time, but these calculations should be close enough.
Mount this high power resistor where it can get a little cooling air. It would be better to get a power rating greater than 20 Watts, as long as the resistance is pretty close to 7.5 Ohms.
You realize that this means that there will be one of these resistors wired in parallel at the point that originally saw the 55 Watt bulb. This means you will need two of these resistors, one for each bulb.
Good luck,
