This is possible iff $n \le 6$. It is relatively easy to write down constructions for $n \le 6$. ConstructionFor $n=1$ take $(1)$. For $n=2$ take $(1),(2)$. For $n=3$ take $(1,3),(2,3)$. For $n=4$ take $(1,3,4),(2,3,4),(4)$. For $n=5$ take $(1,4,5),(2,3,5),(3,4,5),(4,5),(5)$. For $n=6$ take $(1,5,6),(2,5,6),(3,4,6),(3,5,6),(4,5,6),(4,6),(5,6),(6)$. Now let us prove that $n\ge 7$ is impossible. Indeed, of the $f_n$ books written by the $n$-th author, at most $f_1+f_2+\dots+f_{n-3}=f_{n-1}-1$ can have one of the first $n-3$ authors as a coauthor. Hence at least $f_{n-2}+1$ of the books written by the $n$-th author can have only a subset of the last three authors as authors. Since there are at most four such subsets (taking into account that the $n$-th author is always an author of these books), we conclude that $f_{n-2}+1 \le 4$ and hence $f_{n-2} \le 3$ and hence $n \le 6$.