I've never seen the circuit before, but at a first-pass guess it looks like a sort of voltage reference. In a typical resistive divider (replace the FET with a resistor), if the input voltage or the resistance change over time, the output of the divider also changes. In this case, though, the properties of the FET fight that change. It's negative feedback.
So say we're in a steady state. Some current flows through the FET, resulting in a voltage across the FET. That voltage is also on the gate of the FET, causing the FET to turn "on" to some degree. If the FET were all the way on, the voltage across it would be very low, so it would turn itself off again. If it were all the way off, the voltage on the gate would be the full 20V, so it would turn itself on. We're somewhere in between, in an analog mode. The FET goes to some voltage, depending entirely on the transfer characteristics of the FET.
So there's some amount of current flowing, depending on the impedance of the FET in that state. Say something changes to force more current to flow; the resistor warms up and its resistance goes down, or the voltage source rises. More current flows through the FET drain-source. The voltage across the FET drain-source rises. But that means the voltage on the FET gate-source also rises, meaning the impedance of the FET goes down. So more current flows through the FET, but through a lower impedance. So the output voltage won't rise as much as it would if the FET were a resistor.
The inverse applies if something causes less current to flow through the FET. This could be the voltage source falling, the resistance increasing, or a load being applied to the output terminal! If you have a resistive divider, adding a load changes the output voltage. This way, it doesn't, so long as the load isn't so great it eats all the available current from the resistor.
So it's not just a voltage reference, it's a voltage regulator.