Solution:

According to the statement of the problem, we may assume $N>0$ and $d_{1}<d_{2}<\ldots<d_{9}$. Then $k=d_{9}-d_{1}=\max _{1 \leq q<r \leq 9}\left(d_{r}-d_{q}\right)$. Let $z \in \mathbb{N}$ be chosen such that $2^{z}>k$.

Claim: With $N=2^{z} \cdot 3^{z} \cdot 5^{z} \cdot 7^{z} \cdot 11^{z} \cdot 13^{z} \cdot 17^{z} \cdot 19^{z}$ a solution for $N$ has been found.

Proof: We assume that for $n \geq N$ none of the factors of $P(n)$ has a prime factor greater than 20, and consider first $n+d_{1}=2^{e_{1}^{(1)}} \cdot 3^{e_{2}^{(1)}} \cdot \ldots \cdot 19^{e_{8}^{(1)}}$. Since $n \geq N$, there must then be at least one $e_{j}^{(1)}$ with $e_{j}^{(1)} \geq z$ ($1 \leq j \leq 8$). Let the corresponding prime factor be $p$. Thus $n+d_{1}$ is divisible by $p^{e_{j}^{(1)}}$. Since $p^{e_{j}^{(1)}} \geq 2^{\ell_{j}^{(1)}} \geq 2^{z}>k=\max _{1 \leq q<r \leq 9}\left(d_{r}-d_{q}\right)$, none of the other numbers $n+d_{i}$ ($2 \leq i \leq 9$) is divisible by $p^{e_{j}^{(1)}}$, because otherwise $d_{i}-d_{1}$ would be divisible by $p_{j}^{e_{j}^{(1)}}$. Thus $e_{j}^{(1)}=\max _{1 \leq i \leq 9}\left(e_{j}^{(i)}\right)$. Now consider $n+d_{2}$: there, because $d_{2}>d_{1}$, we find at least one $e_{j}^{(2)}$ with $e_{j}^{(2)}>e_{j}^{(1)}$ as well as $e_{j}^{(2)} \geq z$. Hence $n+d_{2}$ also has an $e_{j}^{(2)}$ that is maximal with respect to all factors. This holds correspondingly for the other factors $n+d_{i}$.

Since, however, there are only 8 prime factors less than 20 for the 9 factors of the polynomial, a contradiction arises here by the pigeonhole principle. Therefore a further prime factor must occur, by which $P(n)$ is divisible.