Sum of all numbers whose remainder and quotient are the same
Since this is Code Review, here's a clean-up of your code:

from itertools import count

def sum_numbers_with_equal_quot_rem(divisor):
    result = 0

    for num in count(1):
        if num % divisor == num // divisor:
            result += num

        if num // divisor &gt; divisor:
            break

    return result

sum_numbers_with_equal_quot_rem(40)
#&gt;&gt;&gt; 31980


sys.maxint is the largest int, but not the largest integer, so your loop was broken (although it would have taken a while to find out). This fixes that and makes it Python 3 compatible.

Now, let's add some introspection:

if num % divisor == num // divisor:
    result += num
    print(num % divisor, num // divisor)


and for divisor = 40 we get

1 1
2 2
3 3
4 4
5 5
...
37 37
38 38
39 39


This makes sense as num is fully determined by its divisor and remainder and there must be at least one number with divisor = remainder = k for k in range(1, divisor). There isn't one for 0 or less and there isn't one for divisor or more. This suggest we use:

def sum_numbers_with_equal_quot_rem(divisor):
    result = 0

    for quot_rem in range(1, divisor):
        num = quot_rem * divisor + quot_rem
        result += num

    return result


where num = quot_rem * divisor + quot_rem comes from

help(divmod)
#&gt;&gt;&gt; Help on built-in function divmod in module builtins:
#&gt;&gt;&gt;
#&gt;&gt;&gt; divmod(...)
#&gt;&gt;&gt;     divmod(x, y) -&gt; (div, mod)
#&gt;&gt;&gt;     
#&gt;&gt;&gt;     Return the tuple ((x-x%y)/y, x%y).  Invariant: div*y + mod == x.
#&gt;&gt;&gt;


Namely, the invariant div * y + mod == x.

This gives us an even better option, if we look carefully.

Compare these:

k * divisor + k
(k+1) * divisor + (k+1)


They differ by 1 * divisor + 1. So we are adding up:

$$
1 \, (\text{divisor} + 1) + 2 \, (\text{divisor} + 1) + \cdots + (\text{divisor} - 1)(\text{divisor} + 1)
$$

Letting \$\text{divisor} = d\$, this is just

$$
(d + 1) \sum_{k=1}^{d - 1} k = (d + 1) \left( \frac{1}{2} (d-1) \, d \right) = \frac{d^3 - d}{2}
$$

giving the much simpler

def sum_numbers_with_equal_quot_rem(divisor):
    return (divisor**3 - divisor) // 2

sum_numbers_with_equal_quot_rem(40)
#&gt;&gt;&gt; 31980