Voltage generator and resistor in series
The answer should be easy to see if you look at the complete circuit you created:



The battery is modeled by the 9 V voltage source in series with R1.  R1 represents the internal resistance of the batter, and V1 is its open circuit voltage.  V1 and R1 together form a Thevenin source, which is a better approximation of a battery than just being a perfect voltage source.

R2 is the resistor you put in series with the battery, and R3 is the resistance of your voltmeter.  The meter will show whatever voltage is accross R3.

Now put some plausible values on R1 and R3 (you already said what R2 is).  R1 for a 9 V battery is probably in the 10-20 &Omega; range.  Let's say 20 &Omega;, because as you will see shortly, R1 doesn't really matter in this case.  A typical modern voltmeter will have about 10 M&Omega; resistance, so set R3 to that.

Now you have a voltage divider with R1+R2 forming the top resistor and R3 the bottom.  What is the attenutation ratio of a divider with a top resistor of 240 &Omega; and a bottom resistor of 10 M&Omega;?  From the fact that the bottom resistor is much much larger than the top, you should be able to see this is very nearly 1.  In reality it is so close to 1 that you don't notice the difference on the voltmeter between the top resistor being 20 &Omega; (just the battery) and 240 &Omega; (the battery with your extra resistor in series).