What are the benefits of a high voltage in this LTC6090-based photodiode amplifier?
Why not replace the 10 MΩ resistor with 1 MΩ and get rid of the output
  divider?


It's all about the thermal noise created by the 10 Mohm resistor. A 10 Mohm resistor at 20 degC will produce a noise voltage of 90 uV RMS across a 50 kHz bandwidth. 

Compare this with a 1 Mohm resistor - it produces a noise of 28 uV across the same bandwidth at the same temperature. That isn't a 10:1 reduction in noise voltage and that's the point here.

So, if you use the 10 Mohm resistor (to give ten times the signal gain) then 10:1 attenuate afterwards using "low" value resistors, you will achieve a significant net improvement in signal to noise ratio (\$\sqrt{10}\$). There are other factors that make this not as good as it sounds (such as the noise gain of the circuit due to the photodiode's self capacitance) but, generally speaking, you will get between 5 dB and 10 dB improvment in SNR.

Here's a bit of theory about noise power produced by resistors across a given bandwidth. If you need to see real numbers use this website as a calculator: -



Noise power increases with resistance but, noise voltage only increases with the sqare root of resistance. Signal voltage increases with resistance.


  Why is this circuit improved by a high feedback voltage?


If the target signal output level is (say) 10 Vp-p and you adopt the noise reduction approach shown then you have to produce a signal that is 100 Vp-p in order to be able to attenuate to 10 Vp-p.