4. A natural number $N$ has exactly 12 divisors (including 1 and $N$), and these divisors are numbered in increasing order: $d_{1}<$ $d_{2}<\cdots<d_{12}$. The divisor with the index $d{ }_{4}-1$ equals $\left(d_{1}+d_{2}+d_{1}\right) \times d_{8}$, find $N$.


4. First, use the relationship
$$
d_{d_{4}-1}=\left(d_{1}+d_{2}+d_{4}\right) \times d_{8} \geqslant d_{5} \times d_{8}
$$
$=N$, we get $d_{4}=13, d_{5}=d_{2}+14, \quad N$ $=\left(d_{2}+14\right) \times d_{8}$. Then, using the enumeration method to take $d_{2}=2,3$, $5, 7, 11$ to get the only possible solution $d_{2}=3, d_{3}=9$, thus $N=9 \times 13 \times 17=1989$.