$4 \cdot 10$ natural number $N$ has exactly 12 positive divisors (including 1 and $N$), which are numbered in increasing order: $d_{1}<d_{2}<\cdots<d_{12}$. It is known that the positive divisor with index $d_{4}-1$ equals
$$
\left(d_{1}+d_{2}+d_{4}\right) \cdot d_{8} \text {. }
$$

Find the natural number $N$.
(23rd All-Soviet Union Mathematical Olympiad, 1989)
[Solution] Given that all positive divisors of $N$ satisfy $d_{1}<d_{2}<\cdots<d_{12}$, we know that
$$
d_{1} d_{12}=d_{2} d_{11}=d_{3} d_{10}=d_{4} d_{9}=d_{5} d_{8}=d_{6} d_{7} \text {. }
$$

And $N$ has at most 3 prime divisors.
Let
$$
d_{4}-1=k .
$$

From the problem, we know that $d_{1}+d_{2}+d_{4}$ is a divisor of $d_{k}$,
so $d_{1}+d_{2}+d_{4}$ is also a divisor of $N$, and
$$
d_{1}+d_{2}+d_{4} \geqslant d_{5} \text {. }
$$

Thus, we have $d_{k}=\left(d_{1}+d_{2}+d_{4}\right) \cdot d_{8} \geqslant d_{5} d_{8}=N$.
On the other hand,
$$
d_{k} \leqslant N
$$

Therefore,
$$
d_{k}=N \text {. }
$$

Thus, $k=12, \quad d_{4}=13, \quad d_{5}=d_{1}+d_{2}+d_{4}$.
Since $d_{1}=1$, then $d_{5}=d_{1}+d_{2}+d_{4}=d_{2}+14$.
Since $d_{2}$ is a prime number, and $d_{2} \leqslant d_{4}-2=11$.
We can discuss the cases for $d_{2}=2,3,5,7,11$.
If $d_{2}=2$, then $d_{5}=16=2^{4}$, and $N$'s divisors should also include $2^{2}, 2^{3}$, etc., which is impossible.

If $d_{2}=5$, then $d_{5}=19$, at this point $N$'s divisors already include 3 prime numbers $5,13,19$, so $d_{3}$ should be a composite number, and its divisors should be 5,13, or 19, but for $5<d_{3}<13$ this is impossible.

If $d_{2}=7$, then $d_{5}=21$, at this point $d_{3}$ needs to satisfy $7<d_{3}<13$ and have divisors 3,7, or 13, which is also impossible.

If $d_{2}=11$, then $d_{5}=25$, at this point $d_{3}$ satisfies $11<d_{3}<13$, so $d_{3}=12$, at this time $N$'s divisors should also include $2,3,4,5$, and should be between $d_{1}$ and $d_{2}$, which is impossible.

Therefore, $d_{2}=3$, at this point $d_{5}=17$, and since $N$'s divisors have at most three prime numbers, and $3<d_{3}<13$, thus $d_{3}=3^{2}=9$. At this time,
$$
N=3^{2} \cdot 13 \cdot 17=1989 .
$$

It has exactly 12 positive divisors.