Let's calculate how many numbers between 1 and 30030 are not divisible by any of the numbers from 2 to 16.
Among the numbers from 1 to $n$,

divisible by 2

$$
\frac{n}{2}
$$

divisible by 3 but not by 2

$$
\frac{n}{3}-\frac{n}{6}=\frac{n}{6}
$$

divisible by 5 but not by 2 or 3

$$
\frac{n}{5}-\frac{n}{10}-\frac{n}{30}=\frac{n}{15}
$$

divisible by 7 but not by 2, 3, or 5

$$
\frac{n}{7}-\frac{n}{14}-\frac{n}{42}=\frac{n}{7 \cdot 15}=\frac{4 n}{105}
$$

divisible by 11 but not by 2, 3, 5, or 7

$$
\frac{n}{11}-\frac{n}{22}-\frac{n}{66}-\frac{n}{11 \cdot 15}-\frac{n}{11 \cdot 105}=\frac{8 n}{385}
$$

divisible by 13 but not by 2, 3, 5, 7, or 11

$$
\frac{n}{13}-\frac{n}{26}-\frac{n}{78}-\frac{n}{13 \cdot 15}-\frac{n}{13 \cdot 105}-\frac{8 n}{13 \cdot 385}=\frac{80 n}{5005}
$$

In this case, $n=30030$, so

$$
\begin{gathered}
\frac{n}{2}=15015, \frac{n}{6}=5005, \frac{n}{15}=2002, \frac{n}{105}=1144 \\
\frac{8 n}{385}=624, \frac{80 n}{5005}=480
\end{gathered}
$$

The sum of these numbers is 24270, so $30030-24270=5760$ numbers from 1 to 30030 have the property mentioned in the problem.

(Jánosy Gyula, Budapest.)

The problem was also solved by: Heimlich P., Schuster Gy.