502. Find the smallest fraction, which when divided by each of the fractions $\frac{21}{25}$ and $\frac{14}{15}$, results in a natural number.

The conditions of this problem are unique: here two fractions are mentioned, one of which divides the other exactly.
$\triangle$ Let the required fraction be $\frac{a}{b}$, where $a$ and $b$ are natural coprime numbers. Divide it by each of the given fractions.

$$
\frac{a}{b}: \frac{21}{25}=\frac{25 a}{21 b}, \quad \frac{a}{b}: \frac{14}{15}=\frac{15 a}{14 b}
$$

Since the obtained fractions are equal to natural numbers, $25 a$ is divisible by $21 b$, and $15 a$ is divisible by $14 b$. Considering that the numbers $a$ and $b$ are coprime, we get:

$$
\begin{array}{ll}
a: 21, & 25: b \\
a: 14, & 15: b
\end{array}
$$

Then the number $a$ is a common multiple of the numbers 21 and 14, and the number $b$ is a common divisor of the numbers 25 and 15. But since the required fraction $\frac{a}{b}$ should be the smallest of all possible, its numerator is the smallest, and its denominator is the largest of all possible. Therefore,

$$
a=\text { LCM }(21,14)=42, \quad b=\text { GCD }(25,15)=5
$$

Answer: $\frac{42}{5}$