I believe the answer is $f(n)=\lceil\log[2]{(n+1)}\rceil\ (*)$. For each $i\in\overline{1,n}$, let $v_i$ be the binary vector of length $f(n)$ whose $j$’th component is $1$ iff $i$ belongs to the $j$’th set $A_j$. The hypothesis tells us that these vectors are non-zero and no two are equal, so a necessary and sufficient condition for their existence is $2^{f(n)}-1\ge n$, which gives $(*)$.

grobber wrote: I believe the answer is $f(n)=\lceil\log[2]{(n+1)}\rceil\ (*)$. I am hereby to confirm that your claim is correct, as I am possessing the book 'An Olympiad Down Under' which contains detailed solutions of all the long-listed problems on the 29th IMO. The solution appeared in the book writes: $f(n)= \lceil \log_2 {n} \rceil +1$ if $n=2^r$ and $f(n)= \lceil \log_2 {n} \rceil $ if $n\neq2^r$ Indeed, I have checked that it's the same as yours.

Let $s(n)$ denote the smallest number of seperating subsets of $\{1, \cdots, n\}$. $cs(n)$ denote the smallest number of both seperating and covering subset of $\{1, \cdots, n\}$. We easily get the recursion: a. $s(n) = 1 + min\{max(s(2), s(n-2)), max(s(3), s(n-3)), \cdots \}$ b. $s(n)+1 = cs(n)$. From $a$, we easily get $s(n) = \lceil log_2 n \rceil$, so $cs(n) = s(n) + 1 = \lceil log_2 n \rceil + 1$