14. $[\mathbf{8}]$ Compute
$$
\sum_{k=1}^{2009} k\left(\left\lfloor\frac{2009}{k}\right\rfloor-\left\lfloor\frac{2008}{k}\right\rfloor\right) .
$$
Answer:
$$
2394
$$

Solution: The summand is equal to $k$ if $k$ divides 2009 and 0 otherwise. Thus the sum is equal to the sum of the divisors of 2009 , or 2394 .