We are asked to find the number of ordered arrays $(x_1, x_2, \ldots, x_{100})$ that satisfy the following conditions:
1. $x_1, x_2, \ldots, x_{100} \in \{1, 2, \ldots, 2017\}$,
2. $2017 \mid (x_1 + x_2 + \cdots + x_{100})$,
3. $2017 \mid (x_1^2 + x_2^2 + \cdots + x_{100}^2)$.

To solve this problem, we generalize to an arbitrary prime $p$ and use a classical roots of unity filter to count the number of such tuples. Let $\omega = e^{\frac{2 \pi i }{p}}$ and $N$ be the total number of such ordered tuples.

The key observation is that:
$$\sum_{0 \leq a, b \leq p-1} \omega^{b(x_1 + x_2 + \cdots + x_{100}) + a (x_1^2 + x_2^2 + \cdots + x_{100}^2)} = \begin{cases} p^2 & \text{if } (x_1, x_2, \ldots, x_{100}) \text{ satisfies the conditions}, \\ 0 & \text{otherwise}. \end{cases}$$

From this observation, we see that:
$$p^2 \cdot N = \sum_{(x_1, x_2, \ldots, x_{100})} \sum_{0 \leq a, b \leq p-1} \omega^{b(x_1 + x_2 + \cdots + x_{100}) + a (x_1^2 + x_2^2 + \cdots + x_{100}^2)}.$$

Swapping the sums makes it easier to factor:
$$p^2 N = \sum_{0 \leq a, b \leq p-1} \left( \sum_{x=0}^{p-1} \omega^{ax^2 + bx} \right)^{100}.$$

We deal with the edge case $a = 0$ first. If $b$ is nonzero, then $1 + \omega^b + \omega^{2b} + \cdots + \omega^{(p-1)b} = 0$. On the other hand, if $b = 0$, then the sum evaluates to $p$.

Hence:
$$p^2 N = p^{100} + \sum_{\substack{1 \leq a \leq p-1 \\ 0 \leq b \leq p-1}} \left( \sum_{x=0}^{p-1} \omega^{ax^2 + bx} \right)^{100}.$$

To relate the inner sums to Gauss sums, we complete the square:
$$p^2 N = \sum_{b=0}^{p-1} \sum_{a=1}^{p-1} \omega^{\frac{-b^2}{a}} \left( \sum_{x=0}^{p-1} \omega^{a(x + \frac{b}{2a})^2} \right)^{100}.$$

Since $\omega$ is a primitive $p$th root of unity,
$$\sum_{x=0}^{p-1} \omega^{a(x + \frac{b}{2a})^2} = \sum_{x=0}^{p-1} \omega^{ax^2}.$$

For $a$ not divisible by $p$, define $G(a) = \sum_{x=0}^{p-1} \omega^{ax^2}$ and denote $G(1)$ by $G$. We wish to compute $G(a)$.

Claim: $G(a) = \left(\frac{a}{p}\right) G$.

This follows from the properties of quadratic residues and non-residues. Since $G(a)$ is raised to an even power, its sign does not matter. We need to evaluate $G^{100}$.

Claim: $G^2 = (-1)^{\frac{p-1}{2}} p$.

This follows from the properties of Gauss sums. Using Euler's Criterion, we conclude that $G^2 = (-1)^{\frac{p-1}{2}} p$.

Returning to our expression for $N$:
$$p^2 N = p^{100} + \sum_{b=0}^{p-1} \sum_{a=1}^{p-1} \omega^{\frac{-b^2}{a}} G^{100}.$$

Since the sum over $a$ for fixed $b$ evaluates to zero, we conclude:
$$N = p^{98}.$$

Thus, for $p = 2017$, the number of ordered arrays $(x_1, x_2, \ldots, x_{100})$ that satisfy the given conditions is:
$$\boxed{2017^{98}}.$$