= Solution
The points are $\delta$-spaced if their <circular spacing> satisfies $\|\theta_r-\theta_s\|\geq\delta$ for $r\ne s$, where $\|t\|$ is distance to the nearest <integer>. Ordinary distance on the real line would be insufficient because the <complex exponential> is periodic.
Let $S(t)=\sum_{M<n\leq M+N}a_ne(nt)$ and $F(t)=e(-Mt)S(t)$. Multiplication by this unit-modulus factor leaves $|S(t)|$ unchanged and places the frequencies of $F$ in $1,\ldots,N$. Put $E=\sum|a_n|^2$. The permitted <Sobolev–Gallagher inequality>, in the form needed here, is
$$
|F(t)|^2\leq\delta^{-1}\int_{t-\delta/2}^{t+\delta/2}|F(u)|^2\,du
+\int_{t-\delta/2}^{t+\delta/2}|F(u)F\prime(u)|\,du.
$$
For $0<\delta\leq1$, the arcs about the $\theta_r$ have disjoint interiors on the <circle group>. Summing and applying the <Cauchy-Schwarz inequality> gives
$$
\sum_r|S(\theta_r)|^2\leq\delta^{-1}\int_0^1|F|^2
+\left(\int_0^1|F|^2\right)^{1/2}\left(\int_0^1|F\prime|^2\right)^{1/2}.
$$
The <Cauchy-Schwarz inequality> here follows by expanding $0\leq\int|u-cv|^2$ and minimizing over $c\in\mathbb C$. For completeness, the <finite-interval Parseval identities> follow by expanding the squares: $\int_0^1e(kt)\,dt$ is one at $k=0$ and zero at every other <integer> $k$. Thus $\int|F|^2=E$ and $\int|F\prime|^2\leq4\pi^2N^2E$. We obtain the <exponential-sum large sieve> bound
$$
\boxed{\sum_r|S(\theta_r)|^2\leq(\delta^{-1}+2\pi N)E.}
$$
If $\delta>1$, there is at most one point, and the direct <Cauchy-Schwarz inequality> bound $|S|^2\leq NE$ proves the same assertion.
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