A small town has fewer than 6000 inhabitants. We know that among the children, there are $10 \%$ more girls than boys, and among the adults, there are $15 \%$ more men than women. In the town, there are $20 \%$ more children than adults.

How many people live in the town?
Let $n$ be the number of women living in the city. Then $1.15 n = \frac{23}{20} n$ men, so in total $n + \frac{23}{20} n = \frac{43}{20} n$ adults live in the city. According to the condition, there are 1.2 times as many children as adults. Thus, the number of children is $1.2 \cdot \frac{43}{20} n = \frac{129}{50} n$. We know that there are 1.1 times as many girls as boys, so the total number of children is 2.1 times the number of boys. The number of boys is thus $\frac{129}{50} n : 2.1 = \frac{43}{35} n$, and the number of girls is $\frac{129}{50} n - \frac{43}{35} n = \frac{473}{350} n$. Therefore, the total number of people living in the city is

$$
L = n + \frac{23}{20} n + \frac{43}{35} n + \frac{473}{350} n
$$

None of the fractions in the sum can be simplified, but each term is an integer. Thus, $n$ must be divisible by 20, 35, and 350, which means it must also be divisible by their least common multiple, which is 700.

If $n = 700 \cdot k$, then

$$
L = 700 k + 805 k + 860 k + 946 k = 3311 k
$$

According to the condition, $L < 6000$, so only $k = 1$ is possible. Therefore, the city has 3311 inhabitants, of which 700 are women, 805 are men, 860 are boys, and 946 are girls.