(i) Since $\xi = \frac{-1 + \sqrt{17}}{2}$ is irrational and the polynomial $F x = P x - 2017$ has rational coefficients and $\xi$ as a root, then it will have also the conjugate $\frac{-1 - \sqrt{17}}{2}$ as a root, and therefore it is divided by the polynomial $\varphi x = x^2 + x - 4$. It comes easily from the identity
$$F x = P x - 2017 = x^2 + x - 4 Q x + \kappa x + \lambda,$$
by putting $x = \xi$. Then $\kappa \xi + \lambda = 0$ which gives $\kappa = \lambda = 0$, taking in mind that $\xi$ is irrational. Therefore there exists a polynomial $Q x$ such that:
$$\begin{aligned} F x = P x - 2017 &= x^2 + x - 4 \mathcal{Q} x \\ \Leftrightarrow a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 - 2017 &= x^2 + x - 4 \mathcal{Q} x \end{aligned} \quad (1)$$
From (1) for $x=1$ we get:
$$\begin{aligned} & a_0 + a_1 + \dots + a_n - 2017 = -2Q \quad 1 \\ \Rightarrow \quad a_0 + a_1 + \dots + a_n = 2017 - 2Q \quad 1 \equiv 1 \quad \text{mod } 2 \end{aligned}$$

(ii) We consider the set $a_0, a_1, ..., a_n$ with elements nonnegative integers satisfying the following:
(α) $a_n \xi^n + a_{n-1} \xi^{n-1} + ... + a_1 \xi + a_0 = 2017$ and
(β) the sum $a_0 + a_1 + ... + a_n$ is minimal.
First we observe that: $0 \le a_i \le 3$, for all $i = 1, 2, ..., n-2$.
In fact, if it was not true for someone $i = 1, 2, ..., n-2$, then the elements of the set $a_0, ..., a_{i-1}, a_{i-4}, a_{i-1}+1, a_{i+1}+1, a_{i+2}+1, a_{i+3}, ..., a_n$ would be nonnegative integers, it would satisfy relation (α), while the sum of its elements would be less than of $a_1 + a_2 + ... + a_n$, which is absurd.
Let now $Q x = b_{n-2} x^{n-2} + b_{n-1} x^{n-1} + ... + b_1 x + b_0$. Then from the identity
$a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 - 2017 = x^2 + x - 4 \quad b_{n-2} x^{n-2} + b_{n-3} x^{n-3} + ... + b_1 x + b_0$ we get the equations:
$$\text{a 0 - 2017 = -4b 0 a 1 = -4b 1 + b 0 a 2 = -4b 2 + b 1 + b 0 a 3 = -4b 3 + b 2 + b 1 ........................................ a n-2 = -4b n-2 + b n-3 + b n-4 a n-1 = b n-2 + b n-3 a n = b n-2 a 0 - 2017 = -4b 0 a 1 - b 0 = -4b 1 a 2 - b 1 - b 0 = -4b 2 a 3 - b 2 - b 1 = -4b 3 ........................................ a n-2 - b n-3 - b n-4 = -4b n-2 a n-1 - b n-2 = b n-3 a n = b n-2}$$
In general we have: $a_{i+2} - b_{i+1} - b_i = -4b_{i+2}$, for all $i = 0, 1, ..., n-4$
Since $0 \le a_i \le 3$, for all $i = 1, 2, ..., n-2$, from the first equation we have $a_0 = 1$ and $b_0 = 504$. From the second equation we get $a_1 = 0$ and $b_1 = 126$. From the third equation we get $a_2 = 2$ and $b_2 = 157$. Continuing in the same way we find the sets
$b_0, b_1, b_2, ... b_{14} = 504, 126, 157, 70, 56, 31, 21, 13, 8, 5, 3, 2, 1, 0, 0$
$a_0, a_1, a_2, ... a_{14} = 1, 0, 2, 3, 3, 2, 3, 0, 2, 1, 1, 0, 1, 3, 1$
Therefore the least possible value of the sum $a_0 + a_1 + ... + a_n$ is 23.