1. There are fewer than 500 children on the playground. The percentage of boys among all children equals the number of all girls. How many boys and how many girls are on the playground? Find all possible solutions.
1. Let's denote by $c, d$ the number of boys and girls on the playground, respectively. Then we have

$$
\frac{100 c}{c+d}=d
$$

From this,

$$
c=\frac{d^{2}}{100-d}=\frac{100^{2}-\left(100^{2}-d^{2}\right)}{100-d}=\frac{10000}{100-d}-d-100 .
$$

Since $c$ is a non-negative integer, $100-d$ must be a positive divisor of the number 10000, i.e., the expression $100-d$ can only take the values: $1,2,4,5,8,10,16,20,25,40,50,80$. However, according to the problem, the condition $c+d \leq 16$ must be satisfied. For the values of $100-d$ equal to $20,25,40,50,80$, we get the following solutions step by step:

$$
(c, d)=(320,80),(225,75),(90,60),(50,50),(5,20)
$$

For a complete solution, 6 points are awarded, with 1 point for setting up the equation, up to 3 points for the reasoning leading to the conclusion that the number $100-d$ divides the number 10000, and 2 points for determining all pairs $(c, d)$.