How to determine whether a year is a leap year?

The whole formula can be contained in a single expression:

def is_leap_year(year):
    return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0

print n, " is a leap year" if is_leap_year(n) else " is not a leap year"

Use calendar.isleap:

import calendar
print(calendar.isleap(1900))

You test three different things on n:

n % 4
n % 100
n % 400

For 1900:

1900 % 4 == 0
1900 % 100 == 0
1900 % 400 == 300

So 1900 doesn't enter the if clause because 1900 % 100 != 0 is False

But 1900 also doesn't enter the else clause because 1900 % 4 != 0 is also False

This means that execution reaches the end of your function and doesn't see a return statement, so it returns None.

This rewriting of your function should work, and should return False or True as appropriate for the year number you pass into it. (Note that, as in the other answer, you have to return something rather than print it.)

def leapyr(n):
    if n % 400 == 0:
        return True
    if n % 100 == 0:
        return False
    if n % 4 == 0:
        return True
    return False
print leapyr(1900)

(Algorithm from Wikipedia)

As a one-liner function:

def is_leap_year(year):
    """Determine whether a year is a leap year."""

    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

It's similar to the Mark's answer, but short circuits at the first test (note the parenthesis).

Alternatively, you can use the standard library's calendar.isleap, which has exactly the same implementation:

from calendar import isleap
print(isleap(1900))  # False