Solution:

Consider the fractions $a / b$, where $a$ and $b$ range over integers such that $1 \leq a \leq b \leq n$. We will count these fractions in two ways:

a. By unreduced form. For each denominator $b$, there are $b$ possible numerators $a=1,2, \ldots, b$, so the total number of fractions is
$$1+2+\cdots+n=\frac{n(n+1)}{2}$$

b. By reduced form. Suppose a fraction $a / b$ has been reduced to $c / d$. Given the denominator $d$, there are $\phi(d)$ possible numerators $c$ such that $c \leq d$ and $c / d$ is in lowest terms. To get from $c / d$ back to $a / b$, we must multiply numerator and denominator by a positive integer $k$ such that $k c \leq k d \leq n$. Since $k c \leq k d$ always holds if $c \leq d$, the choice of $k$ is limited only by the inequality $k d \leq n$, which has $\lfloor n / d\rfloor$ solutions. Thus there are $\phi(d)\lfloor n / d\rfloor$ fractions $a / b$ for a given choice of $d$, so the total number of fractions is
$$\sum_{d=1}^{n} \phi(d)\left\lfloor\frac{n}{d}\right\rfloor$$
Since both methods of counting must yield the same answer, the identity follows.