sign unchanged for Dirichlet polynomials? Let $\chi_{q}$ be a primitive Dirichlet character with modulus $q$ (see definition at wikipedia ).
For example for $q=5$ we have
\begin{equation}
\begin{aligned}
\chi_{5,1}&=(1,  1,  1,  1, 0),\\
\chi_{5,2}&=(1,  i, -i, -1, 0),\qquad\qquad \text{(1)}\\
\chi_{5,3}&=(1, -1, -1,  1, 0),\\
\chi_{5,4}&=(1, -i,  i, -1, 0),\\
\end{aligned}
\end{equation}
Let $B(\chi_q)=(1/2)(1-\chi_q(-1))$ so $B(\chi_{5,1})=B(\chi_{5,3})=0,B(\chi_{5,2})=B(\chi_{5,4})=1$. 
Denote $n\in\mathbb{N},j\in\mathbb{N_0}$ and we construct the Dirichlet polynomials $P(n,j,\chi_{q})$
\begin{equation}
\begin{aligned}
P(n,j,\chi_{q})&=\sum_{k=1}^{qn}k^{2j+B(\chi_q)}\chi_q(k)\qquad\qquad \text{(2)}
\end{aligned}
\end{equation}
Proposition A:
$\color{red}{sign(\mathrm{Re}\chi_{q}(-1))}P(n,j,\mathrm{Re}\chi_{q})$ and $\color{red}{sign(\mathrm{Im}\chi_{q}(-2))}P(n,j,\mathrm{Im}\chi_{q})$will remain nonnegative for all $n\geq 1, j\geq 0$.
We use Mathematica numerically verified this proposition for $q=5,7;n=1,5;j=0,...,4$ total 20 cases.
We would like to seek references on a proof of Proposition A. This proposition showed up when we study the Generalized Riemann Hypothesis for L-functions associated primitive Dirichlet character.
We will now provide some background information behind our motivation for asking this question.  This is related to
An Interesting Approach towards proving GRH. For simplicity, we will describe it for odd $\chi$, i.e., $\chi(-1)=-1$. 
Let $\Lambda(s,\chi)$ be the Gamma completed L-function $$\Lambda(s,\chi):=(\pi/q)^{-(s+1)/2}\Gamma((s+1)/2)L(s,\chi)\tag{3}$$ 
where $L(s,\chi)$ denotes
the Dirichlet L-function. It satisfies the functional relation 
$$\Lambda(1-s,\bar{\chi})=\frac{i q^{1/2}}{\tau(\chi)}\Lambda(s,\chi).\tag{4}$$
Let $$\Xi(z)
:=\tfrac12\Lambda(\tfrac12+iz,\chi)+\tfrac12\Lambda(\tfrac12-iz,\bar{\chi}),\tag{5}$$
Notice that $\Xi(x)\in\mathbb{R}(x), x\in\mathbb{R}$.
The Generalized Riemann Hypothesis is equivalent to the statements that
(A) there exists a family of entire function $W_n(z)$ (even in $z$) such that $W_n(z)$ uniformly converges to $\Xi(z)$ in $\{0<\mathrm{Im}z<1/2\}$. Using a corollary
of Hurwitz's theorem in complex analysis we can prove that if all the zeros of $W_n(z)$ are real, then $\Xi(z)$ has no zeros in $\{0<\mathrm{Im} z<1/2\}$;
(B) Find a pair of functions such that $W_n(z)=U_n(z)-V_n(z)$.   $U_n(z),V_n(z)$ are also even in $z$. Prove that all the zeros of $U_n(z),V_n(z)$ are simple and real. 
(C) Prove that the positive zeros of $U_n(z)$ are strictly (say) left-interlacing with those of $V_n(z)$. Using an entire function equivalent to Hermite-Kakeya Theorem for
polynomials,  it can be shown that $W_n(z)$ has only real and simple zeros. 
This approach appeared in a preprint on arXiv (Ref.1) On the zeros of Riemann Xi-function 
Adapting the approach of (Ref.1), we are able to show that the following family of functions $W_n(n,z)$ satisfies condition in statement (A) above
$$
W_n(n,z):=2\omega_n\sum_{j=0}^\infty(-1)^jc_{j}\tilde{S}_{n,2j+1}(\chi)\mathrm{sinc}[\omega_n (z-i\phi_{j})]\\+2\omega_n\sum_{j=0}^\infty(-1)^jc_{j}\tilde{S}_{n,2j+1}(\bar{\chi})\mathrm{sinc}[\omega_n (z+i\phi_{j})].\tag{6}
$$
where
$$
\omega_n:=\tfrac12\log n,\quad
\phi_{n,j}:=2j+\tfrac32,\tag{7}
$$
$$
c_{j}:=\frac{(\pi/q)^j}{\Gamma(j+1)},\qquad\tilde{S}_{n,2j+1}(\chi)=\sum_{k=1}^{n}k^{2j+1}\chi(k)\quad
,\tag{8}\\
$$
Motivation for OP: From (8) we can see, for example, that if the Dirichlet polynomials $\tilde{S}_{5n,2j+1}(\mathrm{Re}\chi_{5,2}), \tilde{S}_{5n,2j+1}(\mathrm{Im}\chi_{5,2})$ have fixed signs for all $n,j$, then we can rewrite (6) as $W_n(z)=U_n(z)-V_n(z)$ to partially satisfy the condition of statement (B).
It remains to prove, following (Ref.1), that $U_n(z),V_n(z)$ have only real zeros and the positive zeros of $U_n(z)$ are strictly interlacing with those of $V_n(z)$.
Update
@GH-from-MO found a counterexample to Proposition A when $q=17,j=0,n=1$. See a related question that I posted at mathoverflow.  Numerical values showed that When $q=17,4\leqslant j\leqslant 9,n=1$ Proposition A holds.  Thus we suspect that Proposition A holds for $q=17,4=j_0\leqslant j,n=1$ and this $j_0$ should depend on $q$.
 A: Here is an incomplete proof that we came up last night. We will only show the proof for odd $\chi_q$, i.e., $\chi_q(-1)=-1$ and thus $B(\chi_q)=1$. To simplify the notation, we will set $u=\mathrm{Re}\chi_q, v=\mathrm{Im}\chi_q$ and $Q(n,2j+1,u)=sign(u(-1))P(n,j,u)=-P(n,j,u)$. 
The goal is to prove $Q(n\geq 1,j\geq 1,u)\geq 0$.
First
\begin{equation}
\begin{aligned}
Q(n,j,u)&=-\sum_{k=1}^{qn}k^{j}u(k)\\
&=-\sum_{k=1}^{q}u(k)\sum_{m=0}^{n}(mq+k)^{j}\\
&=\sum_{m=0}^{n}\color{red}{\sum_{k=1}^{q}(-1)u(k)(mq+k)^{j}}\\
&=:\sum_{m=0}^{n}\color{red}{F(m,j)},\qquad (1)\\
\end{aligned}
\end{equation}
Since for $m\geq 1, j\geq 1$,
\begin{equation}
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}m}F(m,j)
&=jq\sum_{k=1}^{q}(-1)u(k)(mq+k)^{j-1}\\
&=jqF(m,j-1),\qquad (2)\\
\end{aligned}
\end{equation}
$F(m\geq 1,j\geq 1)$ is nondecresing for $m$ if $F(m\geq 1,j-1)\geq 0$.
Thus it is suffice to prove that $F(m\geq 1,0)\geq 0$ and $F(0,j\geq 0)\geq 0$.
\begin{equation}
\begin{aligned}
F(m,j)
&=\sum_{k=1}^{q}(-1)u(k)(mq+k)^{j}\\
&=\sum_{k=1}^{q}(-1)u(k)\sum_{l=0}^j\binom{j}{l}(mq)^l k^{j-l}\\
&=\sum_{l=0}^j\binom{j}{l}(mq)^l\color{red}{\sum_{k=1}^{q}(-1)u(k)k^{j-l}}\\
&=\sum_{l=0}^j\binom{j}{l}(mq)^l\color{red}{F(0,j-l)},\qquad (3)\\
\end{aligned}
\end{equation}
Thus it is suffice to prove that $F(0,j\geq 0)\geq 0$.
For $j\geq 1$,
\begin{equation}
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}j}F(0,j)
&=j\sum_{k=1}^{q}(-1)u(k)k^{j-1}\\
&=jF(0,j-1),\qquad (4)\\
\end{aligned}
\end{equation}
Thus it is suffice to prove that $F(0,1)>0$ because
for primitive and non principal character, we know that $F(0,0)=-\mathrm{Re}\sum_{k=1}^{q}\chi_q(k)=0$.
\begin{equation}
\begin{aligned}
F(0,1)
&=-\sum_{k=1}^{q}ku(k)\\
&=-\sum_{k=1}^{\color{red}{q-1}}ku(k)\\
&=-\left[(q-1)u(q-1)+\cdots+2u(2)+1\right]\\
&=q-2-\sum_{k=2}^{\color{red}{q-2}}ku(k)\quad \because u(q-1)=u(-1)=-1,
\qquad (5)\\
\end{aligned}
\end{equation}
I have no idea how to proceed from here. 
I posted a new question at mathoverflow for proving $F(0,1)$ in (5) is positive.
