It is not a question of measuring temperature, but rather adhering strictly to the datasheet limitations of the SAA7705H DSP chip in the empeg.

The datasheet I have (from 1999, newer ones are google'able) says this:

IO(sink/source) DC output sink or source current:
output type 4mA (BD4CR, BT4CR and B4CR);
-0.5 < VO < VDD + 0.5V: (MAX)±20 mA

So.. I get two possible interpretations:
(1) no single output pin should have to supply more than 4mA, or
(2) each digital output pin has an absolute limit of 20mA.

Let's assume the worst case: no more than 4mA current draw.

In the player I have open here, the scope shows about 6V output on the I2S pins. So (Ohms Law again) 6V / 4mA = 1500ohms minimum between the output and ground.

If all we were connecting to it was the two resistors, that would mean that the lowest value that could be used would be 750+750ohms.

But the middle of the voltage divider is being connected to the WT32i, which itself has a ("internal") resistance. Dunno what that is, but it does draw a small amount of current. The datasheet for it probably says "how much" it draws, but I don't have that within reach where I am, so let's conservatively guess at 1mA draw.

So subtracting that 1mA from the absolute limit of 4mA means the resistor divider thingie can safely pull up to perhaps 3mA from the DSP without frying the output pin. Now feed that into Mr.Ohm and we get 6V/3mA = 2000ohms.

This suggests that 1K/1K is the absolute limit for the two voltage divider resistors. In practice, I would stay above that absolute limit, to allow a bit of margin to not fry anything. So.. perhaps anything larger than 1300ohms might be fine.

Only one formula required for all of this: Ohm's Law: Volts = (Amps * Ohms)

Cheers