I believe that the diode is there to limit the output voltage from the regulator in the event that the -10V source is off or disconnected. Consider power on or off events - you may not have any way of controlling the sequencing of the 35V and -10V supplies.
Under normal operation, the anode voltage of the diode is constant due to the constant current established by the Vref / 120 ohm resistor. With a Vref of 1.25V, there is a current of 10.4mA flowing through R1, R2 and R3. The anode voltage is -10 + 680 * 0.0104 = -2.92V. The diode is therefore reverse-biased and does not have any effect on the circuit.
Now let's consider what happens when the -10V source is disconnected. Normally, the output voltage of the regulator is determined by this equation:
Vref * (1 + (R1 + R3) / R2) - 10 
If the -10V source is disconnected, the regulator output voltage now equals:
Vref * (1 + R1 / R2) + 0.7 
 minus  is:
-Vref * (R3 / R2) + 10.7 
With the established values of Vref, R3 and R2, this yields an overvoltage limit of about 3.6V more than the desired set-point.
Without the diode in the circuit, the output voltage will rise to the regulator maximum whenever the -10V is off or disconnected. This will be about 32V (assuming a 3V regulator dropout). This would definitely be catastrophic if you were using this regulator for example to power a circuit that was only designed for 12V or less.