1. Set up the variables
- Let the number of girls in 2003 be $g$ and the number of boys be $b$.
- Since the number of girls was equal to the number of boys in 2003 , we can say $g=b=x$ .
- The total population in 2003 was $2 x$.
2. Calculate the 2004 population
- In 2004 , both the girl and boy populations increased by $20 \%$.
- Number of girls in $2004=1.2 x$
- Number of boys in $2004=1.2 x$
- Total population in $2004=1.2 x+1.2 x=2.4 x$
3. Apply the integer constraint
- The number of girls ( $1.2 x$ ) and boys ( $1.2 x$ ) in 2004 must be a whole number (an integer).
- $1.2 x$ can be written as $\(\frac{6}{5} x\)$. For this to be an integer, $x$ must be a multiple of 5 .
- The total population ( $2.4 x$ ) can be written as $\(\frac{12}{5} x\)$.
- If $x$ is a multiple of 5 (e.g., $x=5 k$, where $k$ is an integer), then the total population is:

$$
\(\text { Total population }=\frac{12}{5}(5 k)=12 k\)
$$

- Therefore, the total student population in 2004 must be a multiple of $\(\mathbf{1 2 .}\)$
4. Test the options

We check which of the provided options is divisible by 12 (divisible by both 3 and 4):
- 4832: $4+8+3+2=17$ (Not divisible by 3 )
- 5034: Last two digits (34) are not divisible by 4 .
- 5058: Last two digits $(58)$ are not divisible by 4 .
- 5076:
- Sum of digits: $5+0+7+6=18$ (Divisible by 3 )
- Last two digits: $\(76 \div 4=19\)$ (Divisible by 4 )
- $\(5076 \div 12=423\)$. (Multiple of 12 )
- 5128: $5+1+2+8=16$ (Not divisible by 3 )

Conclusion: The only value that could be the total student population in 2004 is 5076.