Start by reducing the circuit to the left of the diode using Thevenin's theorem. The left-hand circuit reduces to a 5V source in series with a 500 ohm resistor. On the right-hand side, convert the 20V source and the 1kohm series resistor to the Norton equivalent of 20mA in parallel with 1kohm. Then combine the two 1kohm resistors in parallel to get 500ohms.
Combining the two current sources on the right-hand side of the diode:
Converting the Norton circuit on the right-hand side to its Thevenin equivalent:
So now we have two 5V sources on each side connected by the 500 ohm resistors, the diode, and switch network.
Since the voltages are equal on both sides, no current will flow between them because they are connected by a passive network.
Therefore there will be no current flowing through the 1N914 diode regardless of the position of the switch.
The link for the 1N914 diode data sheet was added for misdirection. You don't need it, but it is handy to have it on file