= Solution
Put $S(\theta)=\sum_{r=1}^n e(\theta r^3)$. Its modulus is the modulus of the <Fourier transform> in the question, independently of the sign convention. We use <Hua's lemma> in the cubic form $\int_0^1|S|^8\ll_\delta n^{5+\delta}$; the <cubic eighth-moment proof by differencing> establishes this from the <subpower bound for the divisor function>. Therefore
$$
\frac{N^2}{2007}\le\int_m|S|^9\le\sup_{\theta\in m}|S(\theta)|\int_0^1|S|^8
$$
and $n\asymp N^{1/3}$ imply $\sup_m|S|\gg_\delta n^{1-\delta}$. Choose an actual $\theta\in m$ attaining at least half this bound; attainment of the supremum itself is unnecessary.
Fix $\varepsilon>0$ and put $\varepsilon_0=\min(\varepsilon,1/6)$. The <Dirichlet approximation theorem> with $Q=\lfloor N^{1-\varepsilon_0}\rfloor$ gives <coprime integers> $a,q$ satisfying
$$
1\le q\le Q,\qquad \|q\theta\|\le Q^{-1},\qquad |\theta-a/q|\le(qQ)^{-1}\le q^{-2}.
$$
The <cubic Weyl inequality>, which is the quantitative equidistribution estimate needed here, says
$$
|S(\theta)|\ll_\eta n^{1+\eta}\left(q^{-1}+n^{-1}+q/n^3\right)^{1/4}.
$$
Compare it with the lower bound. Choose $\delta,\eta>0$ so small that $b=4(\delta+\eta)/3<\varepsilon_0/2$. It follows that
$$
q^{-1}+n^{-1}+q/n^3\gg_{\delta,\eta}N^{-b}.
$$
But $n^{-1}=O(N^{-1/3})$ and $q/n^3=O(N^{-\varepsilon_0})$. Both are $o(N^{-b})$, so for large $N$ the first term must supply this lower bound. Consequently $q\ll_\varepsilon N^b\ll_\varepsilon N^\varepsilon$, while the approximation above gives $\|q\theta\|\ll N^{\varepsilon_0-1}\le N^{\varepsilon-1}$. Enlarging the constants covers bounded $N$. Thus
$$
\boxed{\theta\in m,\qquad q\ll_\varepsilon N^\varepsilon,\qquad \|q\theta\|\ll_\varepsilon N^{\varepsilon-1}.}
$$
Here $\|\cdot\|$ is <distance modulo one>. A ninth moment of this size therefore forces at least one near-rational frequency, rather than merely a large pointwise sum somewhere outside $m$.
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