Answer:8 numbers. Consider the following arrangement:7,1,9,3,6,2,4,8. This obviously works. Now if there existed an arrangement with 9, it would have both 7 and 5. 7 and 5 can only be next to 1, hence 7 and 5 would have to be at the ends and 1 joined to both, making it a sequence of 3. This is a contradiction. Hence maximum number of cards is 8.

$\boxed{8}$ Notice that there are no cards have numbers divisible by or divide $5$ or $7$. Also, any card can be put next to $1$ because any number is divisible by $1$. So, we need at least two $1$'s such that we put $5$ and $7$ at the ends of the row and next to each is $1$. We must get rid of one of the two numbers $5$ and $7$. Moreover, the sequence $5$, $1$, $8$, $4$, $2$, $6$, $3$, $9$ clearly satisfies the conditions. So, the final answer is $\boxed{8}$.