What difference makes resistance of thermistors?
The standard way to measure a resistive sensor is to make a voltage
divider with the sensor and a fixed resistance between Vcc and GND, like
this:

Vcc
 │
 Rb
 │
 ├─── analog input
 │
 Rx
 │
GND


where Rx is the resistance you want to measure and
Rb is a known bias resistance. The only unknown is the choice
of the bias resistance.

The simple rule of thumb is to choose a bias resistance close to the
typical values of the resistance you want to measure, as this will
optimize your measurement resolution. More specifically, the relative
measurement resolution, which could be stated as the number of ADC steps
you get per percent change in resistance, is maximal when Rx
=&nbsp;Rb, and decreases only slowly as Rx gets
further from Rb.

Now, one could argue that the relative resistance resolution is not
always the best performance metric. You may want to maximize the
absolute resistance resolution instead, or maybe the absolute
conductance resolution. Each of these choices would lead to a different
choice of bias resistor. For the case of a thermistor, however, the
choice is pretty clear: the resistance varies almost exponentially with
temperature, which means that the absolute temperature variations are
practically proportional to the relative resistance variations. If you
want to use temperature resolution as a metric, you could go through a
tedious calculation only to find out that the optimal Rb is
darn close to Rx.

In short, if your thermistor is 1&nbsp;kΩ at 25&nbsp;°C, and you will
use it near 25&nbsp;°C, then a 1&nbsp;kΩ pullup will give you the best
temperature resolution.

Now, measurement resolution may not be the only issue you care about,
and you may want to sacrifice some of it in order to improve something
else. For example:


Current consumption: a 1&nbsp;kΩ thermistor in series with a 1&nbsp;kΩ
pullup will draw 2.5&nbsp;mA (assuming Vcc =&nbsp;5&nbsp;V), which
could rise to near 5&nbsp;mA if it gets really hot. If you are running
off battery power this could be an issue.
Self heating: in that 1&nbsp;kΩ +&nbsp;1&nbsp;kΩ configuration, the
thermistor will dissipate 6.25&nbsp;mW. I've seen the datasheet of a
typical thermistor stating a thermal dissipation constant of
7&nbsp;mW/K. This means the measurement current would heat the
thermistor to 0.9&nbsp;°C above ambient temperature, which you would
need to take into account in your temperature calculations.


Both of these issues would be mitigated by using a larger value for
Rb, only at the cost of measurement resolution. Switching to
a 10&nbsp;kΩ thermistor would mitigate these issues without sacrificing
resolution.