The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\therefore \frac{d y}{d t} \propto y$

$\Rightarrow \frac{d y}{d t}=k y$                              ( $k$ is a constant)

$\Rightarrow \frac{d y}{y}=k d t$

Integrating both sides, we get:

$\log y=k t+C \ldots(1)$

In the year $1999, t=0$ and $y=20000$.

Therefore, we get:

$\log 20000=C \ldots$ (2)

In the year $2004, t=5$ and $y=25000$.

Therefore, we get:

$\log 25000=k \cdot 5+\mathrm{C}$

$\Rightarrow \log 25000=5 k+\log 20000$

$\Rightarrow 5 k=\log \left(\frac{25000}{20000}\right)=\log \left(\frac{5}{4}\right)$

$\Rightarrow k=\frac{1}{5} \log \left(\frac{5}{4}\right)$                     ...(3)

In the year 2009, t = 10 years.

Now, on substituting the values of t, k, and C in equation (1), we get:

$\log y=10 \times \frac{1}{5} \log \left(\frac{5}{4}\right)+\log (20000)$

$\Rightarrow \log y=\log \left[20000 \times\left(\frac{5}{4}\right)^{2}\right]$

$\Rightarrow y=20000 \times \frac{5}{4} \times \frac{5}{4}$

$\Rightarrow y=31250$

Hence, the population of the village in 2009 will be 31250.