Sum of Sum of Divisors

Trivial investigation gives:

$\sum_{n=1}^{100} \tau (n) = \sum_{k=1}^{100} k \left\lfloor\dfrac{100}k\right\rfloor$

Computing this isn't that hard. We'll follow the steps below:

$51\le k\le 100 ~~~\Longrightarrow~\sum_{k=51}^{100}k \left\lfloor\dfrac{100}k\right\rfloor=\sum_{k=51}^{100}k =3775\equiv 775\pmod{1000}$

$34\le k\le 50 ~~~\Longrightarrow~\sum_{k=34}^{50}k \left\lfloor\dfrac{100}k\right\rfloor=2\sum_{k=34}^{50}k =1428\equiv 428\pmod{1000}$

$26\le k\le 33~~~\Longrightarrow~\sum_{k=26}^{33}k \left\lfloor\dfrac{100}k\right\rfloor=3\sum_{k=26}^{33}k =708\equiv 708\pmod{1000}$

$21\le k\le 25 ~~~\Longrightarrow~\sum_{k=21}^{25}k \left\lfloor\dfrac{100}k\right\rfloor=4\sum_{k=21}^{25}k =460\equiv 460\pmod{1000}$

For $1\le k\le 20$ , we can easily calculate it since we are looking for the greatest multiple less than $100$ of each of $1\le k\le 20$ . And the sum is $1928\equiv 928\pmod{1000}$ .

In total, therefore, the answer is:

$775+428+708+460+928=3299\equiv \fbox{299}\pmod{1000}$