Without getting into an explanation of how diodes work electrically, you need to limit the current flowing through the diode to not exceed what it is rated for lest it will burn out.

A typical (red lets say) LED has a 1.8v voltage drop and wants to be drive with 2ma of current (look up the specs on the diodes you wan to use). That means you need to size the resistor so that it flows no more than 2ma. Since the LED voltage drop is constant (for our purposes) the resistor must drop 7.2 (9v - 1.8v) volts.

Ohms law E=IR. 7.2 = .002 R
R = 3600 ohms

Now the only downside to using a battery bigger than 1.8v is that the resistor is going to dissapate some power (as heat). Twinkle, twinkle, little star, power equals i squared r.

P = .002*.002*3600 or 0.0144 watts. Not too much.

Ron's explanation is correct, however, a typical 5mm red LED current is actually 20ma - not 2ma. So, using the same math the current limiting resistor becomes 360 ohms. Standard 10% resistors in that range come in 330 and 390 ohm values - so you'd use a 390 ohm resistor in this case.