Here are a few thoughts about this task:
10KV sine wave = 14.14KV peak (scopes only display peak voltages).
The maximum displayed trace on the scope is 8 vertical divisions of 5V per division = 40V maximum (without using the Y control and any voltmeter display)
Divide this by 2 to get peak to peak
From this it follows that the required minimum attenuation from 10KV sine wave to the input of the scope is, 14,14/20= 707
The minimum probe resistance would need to be 100M Ohm and preferably 1G Ohm. Any less than 100M Ohms will place an inordinate load on the unit under test (UUT) at 14.14KV.
I hope the data and calculations are correct.
I feel sure that a combination of the 2.5KV probe that I mentioned above and some 9M, 10M, 90M, and 100M Ohm resistors will do the job a lot cheaper than $2,000, but high voltage resistors will be required. Theses are big and expensive but dirt cheap compared to the Tek HV probe. These resistors are not usually designed for good frequency response though
I have done quite a bit of HV work, while making a range of scopes and automobile ignition systems, so just a word of caution. If you are rolling your own attenuator, you need to be very careful about the components and insulating materials you use, At 14KV even the wire has to be special. The layout is also critical.
To get a 25KHz bandwidth capacitive peaking will be required across the resistors. This may mean that you will need small 15KV capacitors.
spec
Additional data
The maximum safe voltage on the input of a scope is typically 300V.
The maximum voltage on a 10:1 10M Ohm scope probe is 500V to 600V.