Brazilian Mathematical Olympiad

Let's count the number $N$ of pairs $(j, k)$ such that $j$ divides $k$ and $1 \le j, k \le n$. Fixing $j$, we must count the multiples of $j$, which is $\lfloor \frac{n}{j} \rfloor$. So
$$N = \lfloor \frac{n}{1} \rfloor + \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor + \dots + \lfloor \frac{n}{n} \rfloor$$
Fixing $k$, we must count the number of divisors of $k$, which is $d(k)$ and then $N = d(1) + d(2) + \dots + d(n)$. So
$$d(1) + d(2) + \dots + d(n) = \lfloor \frac{n}{1} \rfloor + \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor + \dots + \lfloor \frac{n}{n} \rfloor$$
and since $x - 1 < \lfloor x \rfloor \le x$
$$\left(\frac{n}{1} - 1\right) + \left(\frac{n}{2} - 1\right) + \dots + \left(\frac{n}{n} - 1\right) < d(1) + d(2) + \dots + d(n) \le \frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{n}$$
$$n \left( \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right) \le d(1) + d(2) + \dots + d(n) \le n \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \right)$$