Example 11 (23rd All-Union Mathematical Olympiad, 1989) A 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 $d_{4}-1$ is equal to $\left(d_{1}+d_{2}+d_{4}\right) \cdot d_{8}$. Find the natural number $N$.
Given that all positive divisors of $N$ satisfy $d_{1}<d_{2}<\cdots<d_{12}$, we know
$$
d_{1} d_{12}=d_{2} d_{11}=d_{3} d_{10}=d_{4} d_{9}=d_{5} d_{8}=d_{6} d_{7} \text {, }
$$

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 $d_{1}+d_{2}+d_{4} \geqslant d_{5}$.
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, $d_{k}=N$.
Thus, $k=12, d_{4}=13, d_{5}=d_{1}+d_{2}+d_{4}$.
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 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
$$
N=3^{2} \cdot 13 \cdot 17=1989 \text {. }
$$

It has exactly 12 positive divisors.