Hello Craig00747
The input pins of an I/O board are either pulled up or pulled down internally to set their rest state. The pull up and pull down resistors are part of the I/O board. Connecting 5v to a pin internally through a resistor means you'll measure 5v at that pin to ground with no external circuitry connected. Connecting the pin internally to ground through a resistor pulls the pin low, 0v, so you measure 0v to ground when you check the pin. The rest state of the I/O pins determines how you configure the external switches or sensors, whether you supply 5v or connect the pin to ground with a limit switch, for example. Connecting a voltage source to a pin that is puled low will flow current through the pull down resistor, and you have to limit that current or the resistor gets hot and burns out. 5v is ok, but 24v needs an external dropping resistor. If all you have are 3.3k then put two in series and try it again. As long as you have more than 3.5v it should work. Never exceed 6.5v or you'll burn out the input chip.
It may help understand the I/O configuration if you have the circuit diagram for the board. Some supply it for you to use.
In my case the board mfr put input led's on the board connected to ground with a 39 ohm resistor. That's not a typical value for a 5v led, and I think they misread the value code on the resistor, 390, which equals 39 ohms, not 390 ohms. 391 equals 390 ohms. Connecting 5v made the led's as bright as highway flares and pulled something like 65ma, way too much. I unsoldered the input led's resistors and now they just have a 220 ohm pull down on the board. Sometimes you have to figure it out on your own.
To reiterate, I do not have the same board that you have and all this is general info that applies to all boards. My sensors have an internal 10k resistor, so my 100 ohm resistors are really not needed.
In very general terms, the input device is feeding voltage to the input pin and then to the pull down resistor. If you know the value of the on-board pull down resistor you can calculate the current flow through it, 5.9v voltage drop. That same current is dropping 18.1 v through the sensor, so you can calculate the total resistance of the sensor plus external resistor. Change the 5.9 to 4.5 and recalculate to find the necessary external resistor. Or use a pot.