ExternalDifferential ADC Values for PIC
For the \$V_r\$ calculation:


\$V_{ref}=\frac{V_{cc}}{2}\$, therefore \$V_{ref}=2.5V\$
the differential input range is from \$-0.5*Vref\$ up to \$0.5*Vref\$, therefore from -1.25...1.25V
the result is in two's complement, with bit 15 as MSB, so the input ranges from -32768 to 32767
(bit 16 acts as additional sign bit, and together with bit 15 acts as overflow / underflow indicator. You might want to check this - look at pages 10+11 in the data sheet)


Try to rewrite the formula as: $$V_r=\frac{ADC_{Meas}}{32768}*1.25+2.5$$
then you see what happens:


first you scale the measured value to the full range
then you multiply it with the full range voltage (1.25V)
then you add the common mode voltage (2.5V)


The same goes for \$R_t\$:


Rt and the 1k resistor form a voltage divider
the formula for a voltage divider is \$\frac{R1}{R2}=\frac{V1}{V1}\$
the voltage across the 1k resistor is \$5V-Vr\$
so the formula is \$\frac{R_t}{R1k}=\frac{V_r}{5-V_r}\$
when you solve this for \$R_t\$, you get the formula above


(update to reflect questions about differential ADCs)

Imagine a differential ADC as two single ADCs sharing a common ground, and a common reference. Both ADC channels of a differential ADC measure the applied input signals with respect to a common ground, and the result is the difference between the two measurements. The typical allowed input range for both channels is between ground and the power supply of the ADC.

In that case, the applied reference voltage as not real relationship to the allowed input signal range. It just affects the differential signal between the two channels. So if you have a differential measurement range of, say, 0.5V, and a supply voltage of 5V, the both signals may range from 4.5 up to 5V (provided the ADC accepts input signals up to Vcc, which the LTC2436 does) without over- or underflow.

The effect is the same as when one input (CH-) is connected to ground while the other one accepts positive and negative inputs. But the differential ADC has no requirement that one signal is connected to ground - it accepts floating input signals, as long as the input stays within certain limits with regard to ground and supply voltage (GND-0.3V up to Vcc+0.3V for the LTC2436-1)

But even then the reference voltage not necessarily corresponds to the measurement range. It might be larger or smaller than that, depending on how the ADC works. E.g. there might be a programmable gain amplifier in the ADC, or the input (or the reference voltage) gets divided by a fixed factor. So changing the reference voltage will change the measurement range, but its not necessarily the same.