XIV OBM

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 $$N=\lfloor \frac{n}{1}\rfloor +\lfloor \frac{n}{2}\rfloor +\lfloor \frac{n}{3}\rfloor +\cdots +\lfloor \frac{n}{n}\rfloor$$ Fixing $k$, we must count the number of divisors of $k$, which is $d(k)$ and then $N=d(1)+d(2)+\cdots +d(n)$. So $$d(1)+d(2)+\cdots +d(n)=\lfloor \frac{n}{1}\rfloor +\lfloor \frac{n}{2}\rfloor +\lfloor \frac{n}{3}\rfloor +\cdots +\lfloor \frac{n}{n}\rfloor$$ and since $x-1<\lfloor x\rfloor \le x$ $$\left(\frac{n}{1}-1\right)+\left(\frac{n}{2}-1\right)+\cdots +\left(\frac{n}{n}-1\right)<d(1)+d(2)+\cdots +d(n)\le \frac{n}{1}+\frac{n}{2}+\cdots +\frac{n}{n}$$ $$n\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)\le d(1)+d(2)+\cdots +d(n)\le n\left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)$$