4. Given that $M$ is the least common multiple of 15 consecutive natural numbers $1,2, \cdots, 15$. If a divisor of $M$ is divisible by exactly 14 of these 15 natural numbers, it is called a "good number" of $M$. Then the number of good numbers of $M$ is $\qquad$.


4.4.

It is known that $M=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$.
Since $2 \times 11, 2 \times 13$ are both greater than 15, therefore,
$$
\begin{array}{l}
\frac{M}{11}=2^{3} \times 3^{2} \times 5 \times 7 \times 13, \\
\frac{M}{13}=2^{3} \times 3^{2} \times 5 \times 7 \times 11,
\end{array}
$$

each is a divisor of $M$, and is exactly divisible by the other 14 numbers.
If the power of 2 in $M$ is reduced by 1, then
$$
\frac{M}{2}=2^{2} \times 3^{2} \times 5 \times 7 \times 11 \times 13
$$

is a divisor of $M$, and is exactly divisible by the 14 numbers except 8.
Similarly, if the power of 3 is reduced by 1, then
$$
\frac{M}{3}=2^{3} \times 3 \times 5 \times 7 \times 11 \times 13
$$

is a divisor of $M$, and is exactly divisible by the 14 numbers except 9;
if the power of 5 in $M$ is reduced by 1, then $\frac{M}{5}$ is a divisor of $M$ and is exactly divisible by the 12 numbers except 5, 10, and 15; if 7 is removed, then $\frac{M}{7}$ is a divisor of $M$ and is exactly divisible by the 13 numbers except 7 and 14.

In summary, there are four divisors of $M$ that are exactly divisible by 14 of the 15 natural numbers, namely $\frac{M}{2}, \frac{M}{3}, \frac{M}{11}, \frac{M}{13}$.