Three, it is known that the natural number $N$ has 20 positive integer factors (including 1 and itself), which are sequentially denoted as $d_{1}, d_{2}, d_{3}, \cdots, d_{21}$, and the factor with index $d_{8}$ is $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$. Find the natural number $N$.
Three, because $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$ is a factor of $N$, so, $d_{1}+d_{4}+d_{6}$ and $d_{4}+d_{8}+d_{13}$ are factors of $N$. Therefore,
$$
\begin{array}{l} 
d_{1}+d_{4}+d_{6} \geqslant d_{7}, d_{4}+d_{8}+d_{13} \geqslant d_{14} . \\
\therefore d_{d 8}=\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right) \\
\geqslant d_{7} \cdot d_{14}=N . \\
\because d_{d 8} \leqslant N, \\
\therefore d_{d 8}=N . \\
\therefore d_{8}=20 . \text{ At this time } d_{1}+d_{4}+d_{6}=d_{7}, d_{4}+d_{8}+ \\
d_{i 3}= d_{14} .
\end{array}
$$

From $d_{8}=20$ we know that $N$ contains the six positive integer factors $1,2,4,5,10,20$, so $N$ must contain the two prime factors 2 and 5.

Since $N$ has 20 positive factors, $20=2 \times 2 \times 5=2 \times 10=4 \times 5$, hence $N$ can be set as $2^{4} \times 5 \times p$ (where $p$ is a prime number not equal to 2 and 5), $2^{9} \times 5$, $2^{3} \times 5^{4}$, or $2^{4} \times 5^{3}$.
(1) When $N=2^{4} \times 5 \times p$,
(1) When $p=3$, $d_{1}, d_{2}, \cdots, d_{8}$ are sequentially $1,2,3,4$, $5,6,8,10$. At this time $d_{8}=10$, which contradicts $d_{8}=20$.
(2) When $p=7$, $d_{1}, d_{2}, \cdots, d_{8}$ are sequentially $1,2,4,5$, $7,8,10,14$. At this time $d_{8}=14$, which contradicts $d_{8}=20$.
(3) When $7<p<23$, $p \geqslant 11$, $d_{1}, d_{2}, \cdots, d_{8}$ are sequentially $1,2,4,5,8,10, p, 16$ or $1,2,4,5,8,10,16, p$, which contradicts $d_{8}=20$.
(4) When $p \geqslant 23$, the positive factors of $N$ are $1,2,4,5,8,10$,
$$
\begin{array}{l}
16,20,40,80, p, 2 p, 4 p, 5 p, 8 p, \cdots . \\
\quad \therefore d_{4}=5, d_{8}=20, d_{13}=4 p, d_{14}=5 p . \\
\therefore 5+20+4 p=5 p .
\end{array}
$$

Thus, $p=25$, which is not a prime number.
Therefore, $N \neq 2^{4} \times 5 p$.
(2) When $N=2^{9} \times 5$, $d_{4}=5, d_{8}=20, d_{13}=128$, $d_{14}=160$. It does not satisfy $d_{4}+d_{8}+d_{13}=d_{14}$.
(3) When $N=2^{3} \times 5^{4}$, $d_{8}=25$, which contradicts $d_{8}=20$.
(4) When $N=2^{4} \times 5^{3}$, $d_{1}=1, d_{4}=5, d_{6}=10$, $d_{7}=16, d_{8}=20, d_{13}=100, d_{14}=125$. Clearly, it satisfies $d_{1}+d_{4}+d_{6}=d_{7}, d_{4}+d_{8}+d_{13}=d_{14}$.
$$
\therefore N=2000 \text{. }
$$

Therefore, the natural number $N$ is 2000.