Solution:

Notice that, for fixed $a$, $\lfloor n / a \rfloor$ counts the number of integers $b \in \{1, 2, \ldots, n\}$ which are divisible by $a$; hence, $f(n)$ counts the number of pairs $(a, b)$, $a, b \in \{1, 2, \ldots, n\}$ with $b$ divisible by $a$. For any fixed $b$, the number of such pairs is $d(b)$ (the number of divisors of $b$), so the total number of pairs $f(n)$ equals $d(1) + d(2) + \cdots + d(n)$. But $d(b)$ is odd precisely when $b$ is a square, so $f(n)$ is odd precisely when there are an odd number of squares in $\{1, 2, \ldots, n\}$. This happens for $1 \leq n < 4; 9 \leq n < 16; \ldots; 81 \leq n < 100$. Adding these up gives 55 values of $n$.