We claim the answer is \[N - \lfloor \frac{N}{4} \rfloor\] which can be attained by taking every element greater than $N/4$ from the original set. If there existed $a, b, c$ with $a | b$ and $b | c$, \[c \ge 2b \ge 2(2a) = 4a\] contradicting the fact that $4a > N$. We must now prove that this is the largest possible $S$. Every number can be written uniquely as $2^lo$ where $o$ is odd and $l$ non-negative. A necessary, but not sufficient, condition is then that there at most two elements of $S$ with the same $o$ value. For \[\lfloor \frac{N}{2} \rfloor + 1 \le o \le N\] (where $l = 0$ by default), we can include at most all these numbers in our set. (The total number of elements here is around $N/4$) For \[\lfloor \frac{N}{4} \rfloor + 1 \le o \le \lfloor \frac{N}{2} \rfloor\] (where $l = 0$ or $l = 1$), we can include at most all these numbers in our set. (The total number of elements here is around $2(N/8) = N/4$) Finally, for \[1 \le o \le \lfloor \frac{N}{4} \rfloor \] we can include only two elements for each value of $o$. (The total number of elements here is about $2(N/8) = N/4$. Combining this with our earlier two totals gives an estimate which is around the $3N/4$ in the answer) Some moderately annoying computation (casework mod 8) gives the desired result.