[proofplan] We prove that every complete block of $q$ consecutive integers contributes zero to the character sum. The key algebraic input is that a nonprincipal character has some reduced residue class $a$ with $\chi(a) \neq 1$, while multiplication by $a$ permutes the reduced residue classes modulo $q$. After decomposing the integers $1 \leq n \leq x$ into complete blocks of length $q$ and one final incomplete block, only the incomplete block remains, and its contribution is bounded by its length. [/proofplan] [step:Show that the sum over one complete residue system modulo $q$ vanishes] Define the complete residue sum $A \in \mathbb{C}$ by \begin{align*} A := \sum_{r=1}^{q} \chi(r). \end{align*} Since $\chi$ is nonprincipal, there exists an integer $a \in \mathbb{Z}$ such that $\gcd(a,q)=1$ and $\chi(a) \neq 1$. Multiplication by $a$ induces a bijection of the residue classes modulo $q$, hence periodicity of $\chi$ gives \begin{align*} A &= \sum_{r=1}^{q} \chi(ar). \end{align*} Using complete multiplicativity of the Dirichlet character, \begin{align*} A &= \sum_{r=1}^{q} \chi(a)\chi(r) = \chi(a) \sum_{r=1}^{q} \chi(r) = \chi(a)A. \end{align*} Therefore $(1-\chi(a))A=0$. Since $\chi(a)\neq 1$, it follows that $A=0$. [guided] We first isolate the cancellation that happens in one full period. Define \begin{align*} A := \sum_{r=1}^{q} \chi(r). \end{align*} Because $\chi$ is nonprincipal, it cannot agree with the principal character on every residue class. Since a Dirichlet character is already $0$ on integers not coprime to $q$, nonprincipality means that there is some integer $a \in \mathbb{Z}$ with $\gcd(a,q)=1$ and $\chi(a)\neq 1$. Now multiplication by $a$ is invertible modulo $q$, because $\gcd(a,q)=1$. Hence the map on residue classes modulo $q$ given by \begin{align*} r \pmod q \mapsto ar \pmod q \end{align*} is a bijection. Since $\chi$ is periodic modulo $q$, replacing the complete list of residues $1,\dots,q$ by the complete list $a,2a,\dots,qa$ modulo $q$ does not change the sum: \begin{align*} A &= \sum_{r=1}^{q} \chi(ar). \end{align*} By complete multiplicativity of $\chi$, \begin{align*} \sum_{r=1}^{q} \chi(ar) &= \sum_{r=1}^{q} \chi(a)\chi(r) = \chi(a)\sum_{r=1}^{q}\chi(r) = \chi(a)A. \end{align*} Thus $A=\chi(a)A$, so \begin{align*} (1-\chi(a))A=0. \end{align*} The coefficient $1-\chi(a)$ is nonzero by the choice of $a$, and therefore $A=0$. [/guided] [/step] [step:Decompose the partial sum into complete blocks and one remainder] Let $x \geq 1$ be fixed, and define $N := \lfloor x \rfloor \in \mathbb{N}$. Choose integers $M \geq 0$ and $R \in \{0,1,\dots,q-1\}$ by Euclidean division: \begin{align*} N = Mq + R. \end{align*} Then \begin{align*} \sum_{1 \leq n \leq x}\chi(n) &= \sum_{n=1}^{N}\chi(n) \\ &= \sum_{m=0}^{M-1}\sum_{r=1}^{q}\chi(mq+r) + \sum_{r=1}^{R}\chi(Mq+r). \end{align*} For each $m \in \{0,\dots,M-1\}$, periodicity modulo $q$ gives $\chi(mq+r)=\chi(r)$ for every $r \in \{1,\dots,q\}$, so each complete block has sum $A=0$. Hence \begin{align*} \sum_{1 \leq n \leq x}\chi(n) = \sum_{r=1}^{R}\chi(Mq+r). \end{align*} [/step] [step:Bound the remaining incomplete block uniformly in $x$] For every integer $n$, the Dirichlet character value satisfies $|\chi(n)| \leq 1$. Therefore \begin{align*} \left|\sum_{1 \leq n \leq x}\chi(n)\right| &= \left|\sum_{r=1}^{R}\chi(Mq+r)\right| \\ &\leq \sum_{r=1}^{R}|\chi(Mq+r)| \\ &\leq R \\ &\leq q-1. \end{align*} Thus the theorem holds with \begin{align*} C_q := q. \end{align*} This constant depends only on $q$ and satisfies the required bound for every $x \geq 1$. [/step]