How can I wire a 9-position switch so that each position turns on one more LED than the one before?

Using a regulated current source to light them, wire the LEDs in series and short out the segment which you want to be dark.

^{simulate this circuit – Schematic created using CircuitLab}

You can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.

Here's a simple way to build one using a LM2596S module:

1. Remove the potentiometer and both large capacitors
2. Connect one of the salvaged capacitors between +in and +out (positive to +in), and fit a 1uF ceramic capacitor where the output capacitor was.
3. Connect a 100 ohm resistor from the -output to the centre potentiometer terminal.

Modified in this way, it will create a negative voltage on the -out terminals and act as a 12.5mA current sink at the centre potentiometer terminal (with source at +out) if power is applied between +in and +out.

^{simulate this circuit}

or a XL6009 buck-boost module can be modified. this time just remove the potentiometer and add a 100 ohm resistor, connec 3-30V to the nirmal input terminals and connect the LED string to the output and resistor.

^{simulate this circuit}

One way to achieve your progressive LED lightup as you turn the rotary switch is to use a current sink on the common of the switch and then wire the LEDs across the selector switch terminals as shown below. The constant current sink shown is a low cost way to get a 20mA sink for the LEDs so that there is no brightness variation as the number of lit LEDs changes. This scheme does require a high enough supply voltage that overcomes the forward voltage drop of the series string of up to nine LEDs.

If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.