Five. (16 points) A certain middle school originally had several classrooms, each with an equal number of desks, totaling 539 desks. This year, the school added 9 new classrooms in a newly built teaching building, increasing the total number of desks to 1080. At this point, the number of desks in each classroom remained equal, and the number of desks in each classroom increased compared to before. How many classrooms are there now?
Five, let there be $x$ classrooms now, then the original number of classrooms is $x-9$. According to the problem, we have
$\frac{1080}{x}>\frac{539}{x-9}$, which means $\frac{1080}{x}-\frac{539}{x-9}>0$,
or $\frac{1080 x-9720-539 x}{x(x-9)}>0, x>\frac{9720}{541}$.
Thus, $x>17$.
Also, since $\frac{539}{x-9}$ is a natural number, and 539 is an odd number,
$x-9$ must be an odd number.
Therefore, $x$ is an even number. Hence, the possible values of $x$ are even numbers greater than or equal to 18, and $\frac{1080}{x}$ and $\frac{539}{x-9}$ are natural numbers.
When $x=18$, $\frac{539}{18-9}$ is not a natural number, so it is discarded.
When $x=20$, $\frac{1080}{20}=54, \frac{539}{20-9}=49$.
Thus, $x=20$ is a solution that satisfies the conditions.
Since $1080=2^{3} \cdot 3^{3} \cdot 5$,
the even factors of 1080 greater than 20 and less than itself are 30, 40, 60, 120, 360, 540. And
$$
\begin{aligned}
30-9= & =21,40-9=31,60-9=51,120-9= \\
111,360-9 & =351,540-9=531
\end{aligned}
$$

None of these can divide 539, so these even numbers do not satisfy the conditions.
Therefore, $x=20$ is the only solution that satisfies the conditions.
Answer: The school now has 20 classrooms.
(Author: Taiyuan Education College, 030001,