Notice that $2 k$ is a divisor of 1000 iff $k$ is a divisor of 500, so we need only find the sum of the divisors of 500 and multiply by 2. This can be done by enumerating the divisors individually, or simply by using the formula: \sigma\left(2^{2} \cdot 5^{3}\right)=\left(1+2+2^{2}\right)(1+5+5^{2}+5^{3}\right)=1092, and then doubling gives 2184. Alternate Solution: The sum of all the divisors of 1000 is $\left(1+2+2^{2}+2^{3}\right)\left(1+5+5^{2}+5^{3}\right)=2340$. The odd divisors of 1000 are simply the divisors of 125, whose sum is $1+5+5^{2}+5^{3}=156$; subtracting this from 2340, we are left with the sum of the even divisors of 1000, which is 2184.