55 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 $n$.