17. A.

Let $n=\frac{N}{5}$. Arrange $N$ green balls in a row, dividing them into five equal parts. Insert a red ball randomly into one of the $N+1$ gaps, which gives $N+1$ different ways to insert the red ball. When the red ball is placed in the gaps within the middle part, it does not meet the condition that the green balls on one side of the red ball account for at least $\frac{3}{5}$ of the total green balls. Inserting the red ball into the other four parts satisfies the condition. The middle part has $n$ green balls, which creates $n-1$ gaps, so
$$
\begin{array}{l}
P(N)=1-\frac{n-1}{N+1}=1-\frac{n-1}{5 n+1}=\frac{4 n+2}{5 n+1} . \\
\text { When } P(N)=\frac{4 n+2}{5 n+1}=479 .
\end{array}
$$

Thus, the smallest positive integer $N=480$. At this point, $4+8+0=12$.
17. Let $N$ be a positive integer, and $5 \mid N$. A red ball and $N$ identical green balls are randomly arranged in a row. Let $P(N)$ be the probability that the number of green balls on one side of the red ball is at least $\frac{3}{5}$ of the total number of green balls. It is not hard to see that $P(5)=1$, and as $N$ approaches infinity, $P(N)$ is approximately $\frac{4}{5}$. Then the smallest positive integer $N$ such that $P(N) << \frac{321}{400}$ has a sum of its digits equal to ( ).
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20