mini Christmas lights
Yes you can - but a simplistic attempt will lead you astray.
Other advice given so far may mislead (and/or kill) you.  

ALL these questions are important:


Does it use a low voltage power supply?
Some don't!!!
If it is direct mains connected - as some such are - then it can easily kill you.  
Can you supply a well focused photo of the light string and of the power supply spec plate (if it has one).
If it has a power supply, do you know the power supply voltage?
Is it AC or DC ?
(Both should be on the power supply).
How many wires from the string to the power input? 
MAY be 2. May be more. 
What is your mains voltage


If it is direct mains plug in you can still do what you want but need to take exceptional care.

Using AC or DC range to suit supply, using an appropriate range use a DMM (digital multimeter) or similar to measure the voltage across the entire operating string.
Result = ?

Using AC or DC range to suit supply, using an appropriate range use a DMM (digital multimeter) or similar to measure the voltage across a single operating bulb.
Result = ?

If Vbulb ~= Vstring / 20 then they are in series.  

If direct mains connected Vbulb is
= 110/20 ~~= 5.5VAc or
230 VAC/20 ~~= 11 VAC   

Set DMM to AC or DC current range.
Insert DMM in series with whole string.
DO NOT KILL YOURSELF :-( :-)
Measure string current.
Result = ?



Here is what you need to do for the worst case - and most likely - direct mains connection.
If your string is not connected this way, provide the test results connected and I/we will advise further.

If bulbs are in series and mains powered directly.  

Rseries = V/I = Vstring_measured (in volts) / I string measured (in AMPS)

eg if Vstraing = 110 VAC, and I string = 150 mA = 0.15 AMP
Rseries = 110/0.15 =~ 33,000 Ohms.

Ideally use 2 resistors of half the value each in series. This redcues the voltage across each - some resistors are NOT rated for mains voltage use, regardless of power 

Resistor power rating = Vmeasured x I measured x 2
In this example = 110 x 0.15 x 2 = 33 Watts.
That is a larger value than you'd hope - but 3 x 10,000 Ohm x 10 Watt resistors in series will do the job.  

Note I have replaced effectively 20 bulbs - not just 18 - this will run the string VERY SLIGHTLY lower in output than before. You will not notice the difference.