If the output resistance is a real problem, you could connect an external semi-precision power supply through an appropriate resistor to the K-V divider input and adjust the external supply for a null reading at the terminals against your 732A output. That would greatly decrease the loading on the 732A.
This type of measurement has problems, to say the least. You must have a power supply with noise and drift under 1 microvolt at 10 volts nominal. Furthermore it must have a resolution of at least 1PPM. I find these circuits pick up a lot of noise. I have not found a way to use this type of circuit in my lab.
This suggestion is not so critical as you say, but does require a decent power supply and maybe a G-R resistor box. The idea is to first adjust either the supply or the resistor for null against the 732A output, then remove the null meter and leave the power supply (maybe 20 V or so) and resistor (100 k for a 100 k K-V) connected, along with the 732A that will determine the 10 V into the K-V divider. Any error or drift in the current from the supply through the resistor will react with the small-ish output resistance of the 732, not the high-ish resistance of the K-V.
With round numbers, if the 732A has an output resistance of 0.2 ohms, and if the external power supply is set to 20 V through a 100 kohm resistor to the 100 kohm input of the K-V divider, then to get an error of 1 microvolt (out of 10 V at the K-V input) would require an error (or drift) of 5 microamps through the external resistor, which would be 5% of the 100 microamp current of 10 V across 100 kohms, where 5 microamps X 0.2 ohms = 1 microvolt.