Transistor Voltage Inverter
If you pull up the voltage through the 1K to 20V, which is what I assume you mean, then the threshold will still be at about 2V. 

It is set by the Vbe of Q1 for Q1 'on', which is about 0.6V to 0.7V. So the input voltage will be Vbe * (10K+4.7K)/4.7K = 3.1*Vbe = 1.9~2.2V. 

One consideration might be that the 1K resistor will see much more dissipation. So the schematic may be the same, however the 1K should be rated at 1/2W or better (it sees 400mW dissipation worst-case with the switch closed), whereas it only saw 144mW at 12V. This is because the dissipation increases with voltage squared.