39 The numbers $1,2, \cdots, 33$ are written on the blackboard. Each time, it is allowed to perform the following operation: take any two numbers $x, y$ from the blackboard that satisfy $x \mid y$, remove them from the blackboard, and write the number $\frac{y}{x}$. Continue until there are no such two numbers on the blackboard. How many numbers will be left on the blackboard at least?
39. Consider the objective function $S=$ the product of all numbers on the blackboard.

Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $k$ results in $S$ becoming $\frac{S}{xy} \cdot k = \frac{S}{x^2}$. This indicates that after each operation, the parity of the exponent of each prime factor in $S$ remains unchanged. Specifically, $2, 3, 5,$ and $11$ always divide the $S$ obtained after each operation. Since $2 \times 3 \times 5 \times 11 > 33$, at least two numbers must remain on the blackboard such that their product is a multiple of $2 \times 3 \times 5 \times 11$.

Furthermore, note that the primes $17, 19, 23, 29, 31$ have no multiples greater than themselves that are less than or equal to 33. Therefore, no operation can remove any of these numbers.

The above discussion shows that at least 7 numbers must remain on the blackboard.
The following example demonstrates that exactly 7 numbers can remain:
$$\begin{array}{l}
(32,16) \rightarrow 2, (30,15) \rightarrow 2, (28,14) \rightarrow 2, (26,13) \rightarrow 2, (24,12) \rightarrow 2 \\
(22,11) \rightarrow 2; (27,9) \rightarrow 3, (21,7) \rightarrow 3, (18,6) \rightarrow 3; (25,5) \rightarrow 5 \\
(20,4) \rightarrow 5; (8,2) \rightarrow 4 \\
(5,5) \rightarrow 1; (4,2) \rightarrow 2; (3,3) \rightarrow 1, (3,3) \rightarrow 1, (2,2) \rightarrow 1, (2,2) \rightarrow 1 \\
(2,2) \rightarrow 1
\end{array}$$

Thus, the blackboard is left with $10, 17, 19, 23, 29, 31, 33$ and 7 ones. The 7 ones can be removed by pairing them with 17 in 7 operations.

In conclusion, at least 7 numbers remain.