Interpret the $x_i$ as residue classes modulo $2017$. Let $\omega = e^{2 \pi i / 2017}$. Say that a $100$-tuple $(x_1, \dots, x_{100})$ is smooth if it satisfies the conditions. Observe that \[\sum_{a, b} \omega^{a(x_1^2 + \dots + x_{100}^2) + b(x_1 + \dots + x_{100})} = \begin{cases*} 2017^2 & if $(x_1, \dots, x_{100})$ is smooth\\ 0 & otherwise, \end{cases*}\]where $a$ and $b$ range over all residue classes modulo $2017$. Thus, the desired count is equal to \begin{eqnarray*} & & \frac{1}{2017^2} \sum_{a, b} \sum_{x_1, \dots, x_{100}} \omega^{a(x_1^2 + \dots + x_{100}^2) + b(x_1 + \dots + x_{100})}\\ & = & \frac{1}{2017^2} \sum_{a, b} \left(\sum_{x} \omega^{ax^2 + bx}\right)^{100}, \end{eqnarray*}where $x$ ranges over all residues modulo $2017$. Thus, for each $a$ and $b$, it suffices to find $\left(\sum_{x} \omega^{ax^2 + bx}\right)^{100}$. First, we deal with the case $a = 0$. In this case, we wish to find $\left(\sum_{x} \omega^{bx}\right)^{100}$, which is $2017^{100}$ if $b = 0$ and $0$ otherwise. Now, suppose $a \neq 0$, and let $s = \sum_{x} \omega^{ax^2 + bx}$; this becomes \[s = \sum_{x} \omega^{ax^2 + bx} = \omega^{-\frac{b^2}{4a}} \sum_{x} \omega^{a(x + \frac{b}{2a})^2} = \omega^{-\frac{b^2}{4a}} \sum_{x} \omega^{ax^2}.\] Now, we see that \[s^2 = \omega^{-\frac{b^2}{2a}} \sum_{x, y} \omega^{a(x^2 + y^2)}.\] Claim. For each residue $t$ modulo $2017$, the number of solutions to $x^2 + y^2 = t$ is $4033$ if $t = 0$ and $2016$ otherwise. Proof. Well-known; will edit in later if requested. Thus, we see that $\sum_{x, y} \omega^{a(x^2 + y^2)} = 4033 + \sum_{x \neq 0} 2016 \omega^x = 2017$, so $\sum_{x} \omega^{ax^2} = \pm \sqrt{2017}$. This means that \[\left(\sum_{x} \omega^{ax^2 + bx}\right)^{100} = \left(\omega^{-\frac{b^2}{4a}} \sum_{x} \omega^{ax^2}\right)^{100} = \omega^{-\frac{25b^2}{a}} 2017^{50}.\]Thus, we see that \begin{eqnarray*} & & \frac{1}{2017^2} \sum_{a, b} \left(\sum_{x} \omega^{ax^2 + bx}\right)^{100}\\ & = & \frac{1}{2017^2} \sum_{b} \left(\sum_{x} \omega^{bx}\right)^{100} + \frac{1}{2017^2} \sum_{\substack{a, b \\ a \neq 0}} \left(\sum_{x} \omega^{ax^2 + bx}\right)^{100}\\ & = & \frac{2017^{100}}{2017^2} + \frac{1}{2017^2} \sum_{\substack{a, b \\ a \neq 0}} \omega^{-\frac{25b^2}{a}} 2017^{50}\\ & = & 2017^{98} + 2017^{48} \sum_{\substack{a, b \\ a \neq 0}} \omega^{ab^2}\\ & = & 2017^{98} + 2017^{48} \sum_{a \neq 0} 1 + 2017^{48} \sum_{\substack{a, b \\ a \neq 0, b \neq 0}} \omega^{ab^2}\\ & = & 2017^{98} + 2017^{48} \cdot 2016 + 2017^{48} \sum_{b \neq 0} -1\\ & = & 2017^{98}, \end{eqnarray*}so the number of smooth $100$-tuples is $2017^{98}$.

We'll generalise to an arbitrary prime $p$. This is simply the case $p=2017$. Say that an ordered tuple that satisfies the above three conditions is "cool."We shall use a classical roots of unity filter to count the number of cool-tuples. Let $\omega = e^{\frac{2 \pi i }{p}}$ and $N$ be the total number of cool ordered tuples. The key observation is that \[ \sum_{(0 \leq a,b \leq p-1} \omega^{b(x_1+x_2+\dots+x_{100}) + a (x_1^2 + x_2^2 + \dots + x_{100}^2 )} = \begin{cases} p^2 &\text{ if $(x_1, x_2, \dots x_{100})$ is cool } \\ 0 &\text{ otherwise.} \end{cases} \]This follows from summing the terms of a geometric sequence. From the above observation, we see that \[ p^2 \cdot N = \sum_{(x_1, x_2, \dots, x_n)} \sum_{0 \leq a,b \leq p-1} \omega^{b(x_1+x_2+\dots+x_{100}) + a (x_1^2 + x_2^2 + \dots + x_{100}^2 )} .\]Swapping sums makes it easier to factor: \[ p^2N = \sum_{0 \leq a,b, \leq p-1} \left( \sum_{x=0}^{p-1} \omega^{ax^2+bx} \right)^{100} .\]We deal with the edge case $a=0$ first. If $b$ is nonzero, then $1+\omega^b+\omega^{2b}+\dots+\omega^{(p-1)b}=0$. On the other hand if $b=0$, then the sum evaluates to $p$. Hence \[ p^2 N = p^{100} + \sum_{\substack{ 1 \leq a \leq p-1 \\ 0 \leq b \leq p-1}} \left( \sum_{x=0}^{p-1} \omega^{ax^2+bx} \right)^{100} . \]In an effort to relate the inner sums to Gauss sums, we complete the square. \[ p^2N = \sum_{b=0}^{p-1} \sum_{a=1}^{p-1}\omega^{\frac{-25b^2}{a^2}} \left(\sum_{x=0}^{p-1} \omega^{a(x+\frac{b}{2a})^2} \right)^{100}. \]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 for brevity's sake denote $G(1)$ by $G$.We wish to compute $G(a)$. Claim: $G(a) = \left(\frac{a}{p}\right) G$. Merely split into cases for proof. First, that $\left(\frac{a}{p}\right)=1.$Then $(ax^2)_{x=0}^{p-1}$ is merely a permutation of $(x^2)_{x=0}^{p-1}$ so $G(a)=G$. Secondly, we consider the case that $\left(\frac{a}{p}\right) = -1$.Then $G(a)$ runs over zero and each of the quadratic non-residues twice. Note \begin{align*} G(a) &= 2\sum_{\left(\frac{x}{p} \right)= -1} \omega^ x \\ &= 1 + 2 \left( \sum_{i=1}^{p-1} \omega^i - \sum_{\left(\frac{x}{p} \right) = 1} \omega^x \right) \\ &= -G. \end{align*}This concludes the proof of the claim. Recall that the $G(a)$ is raised to an even power so its sign does not matter. What is left is to evaluate $G^{100}$. Claim: $G^2 = (-1)^{\frac{p-1}{2}} p$. First note that $\overline{G}=\left(\frac{-1}{p}\right)G$. For proof, again consider the two possible cases based on the value of the Legendre symbol. If -1 is a quadratic residue than $-x^2$ is also a quadratic residue by multiplicativity so $(-x^2)_{x=0}^{p-1}$ permutes $(x^2)_{x=0}^{p-1}$. The other case cycles over zero and the non-residues, which we've already shown to be equal to $-G$ earlier. Now consider the product \begin{align*} \left(\frac{-1}{p}\right) G^2 &= G \cdot \overline{G} \\ &= \left( \sum_{x=0} \omega^{x^2} \right) \left( \sum_{y=0}^{p-1} \omega^{-y^2} \right) \\ &= \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \omega^{x^2-y^2} \\ &= \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \omega^{(x+y)(x-y)} . \end{align*}There is a bijection between ordered pairs $(x,y)$ and ordered pairs $(u,v)$ such that \[ \begin{cases} x+y = u \\ x-y = v \end{cases} \]Hence \[ \left( \frac{-1}{p}\right) G^2 = \sum_{u=0}^{p-1} \sum_{v=0}^{p-1} \omega^{uv}. \]If $u$ is nonzero, then $\sum_{v=0}^{p-1} \omega^{uv}=0$ and otherwise it is $p$. It follows that the total sum is $p$. With an application of Euler's Criterion, we conclude that $G^2=(-1)^{\frac{p-1}{2}}p$. Coming back to our expression for $N$, \begin{align*} p^2 N &= p^{100} + \sum_{b=0}^{p-1} \sum_{a=1}^{p-1} \omega^{\frac{-25b^2}{a}} G^{100} \\ &= p^{100} + (p-1)p^{50} \sum_{b=0}^{p-1} \sum_{a=1}^{p-1} \omega^{\frac{-25b^2}{a}} . \end{align*}We need to evaluate that pesky last sum. Like before, observe that for any fixed $b$ \begin{align*} \sum_{a=1}^{p-1} \omega^{\frac{-25b^2}{a}} &= \sum_{a=1}^{p-1} \omega^{ab^2}\\ &=\sum_{a=1}^{p-1}G(a)\\ &= \sum_{a=1}^{p-1}\left(\frac{a}{p}\right)G. \\ &= 0 \end{align*}as there are precisely $\frac{p-1}{2}$ quadratic nonresidues and the same number of nonzero quadratic residues. We conclude $N = p^{98}.$