Project Euler Problem 1
Python hint number 1:

The pythonic way to do :

n = 1
while n &lt; 1000:
    # something using n
    n = n + 1


is :

for n in range(1,1000):
    # something using n


Python hint number 2:

You could make your code a one-liner by using list comprehension/generators :

print sum(n for n in range(1,1000) if (n%3==0 or n%5==0))


Your code works fine but if instead of 1000, it was a much bigger number, the computation would take much longer. A bit of math would make this more more efficient.

Math hint number 1 :

The sum of all the multiples of 3 or 5 below 1000 is really the sum of (the sum of all the multiples of 3 below 1000) plus (the sum of all the multiples of 5 below 1000) minus the numbers you've counted twice.

Math hint number 2 :

The number you've counted twice are the multiple of 15.

Math hint number 3 :

The sum of the multiple of 3 (or 5 or 15) below 1000 is the sum of an arithmetic progression.