Solution:
The $i$th term on the left is the sum of all $d$ dividing $i$. If we write this sum out explicitly, then each term $d=1,2,\ldots,n$ appears $\lfloor n/d \rfloor$ times—once for each multiple of $d$ that is $\leq n$. Thus, the sum equals
$$\begin{aligned} \lfloor n/1 \rfloor + 2\lfloor n/2 \rfloor + 3\lfloor n/3 \rfloor + \cdots + n\lfloor n/n \rfloor &\leq n/1 + 2n/2 + 3n/3 + \cdots + n/n \\ &= n + n + \cdots + n \\ &= n^{2}. \end{aligned}$$