= Solution
Write $e(t)=e^{2\pi it}$ and $N=n+1$. A convenient form of the <quadratic Weyl inequality> is the following: if $q\ge1$, $(a,q)=1$ and $|\alpha-a/q|\le q^{-2}$, then
$$
\boxed{\left|\sum_{x=0}^{N-1}e(\alpha x^2)\right|^2
\ll\left(\frac{N^2}{q}+N+q\right)\log(2q).}
$$
The implied constant is absolute. In particular the familiar square-root saving is $|S|\ll(N^2/q+N+q)^{1/2}\log(2q)^{1/2}$. We prove all estimates used in this statement.
First, the finite <geometric series> identity gives
$$
\left|\sum_{x=0}^{L-1}e(\theta x)\right|
\le\min\left(L,\frac1{2\|\theta\|}\right),
$$
where $\|\theta\|$ denotes distance to the nearest integer and the second bound is interpreted as infinite when that distance is zero. Indeed the sum is $(1-e(L\theta))/(1-e(\theta))$, and $|1-e(\theta)|=2|\sin\pi\theta|\ge4\|\theta\|$. This proves the <exponential geometric sum bound>.
Expand the squared <quadratic exponential sum> by the difference $h$ between its indices. Since $(x+h)^2-x^2=2hx+h^2$, the diagonal and the paired off-diagonals give
$$
|S|^2\le N+2\sum_{h=1}^{N-1}\min\left(N,\frac1{2\|2\alpha h\|}\right).
$$
For $q\ge16$, partition the $h$ into consecutive blocks of length at most $L=\lfloor q/8\rfloor$. Within one block, two distinct indices have difference $0<|d|<q/2$. Since the denominator of $2a/q$ after reduction is at least $q/2$, $\|2ad/q\|\ge1/q$. Therefore
$$
\|2\alpha d\|\ge\|2ad/q\|-2|d|\,|\alpha-a/q|
\ge\frac{3}{4q}.
$$
Thus the points $2\alpha h$ in a block are separated on the circle by at least $3/(4q)$.
For any such separated set, at most an absolute number of points lie in each pair of arcs whose distances from zero are between $j/(4q)$ and $(j+1)/(4q)$. The closest few contribute $O(N)$ to the capped reciprocal sum; the remaining arcs contribute $O(q/j)$ for $1\le j\le4q$. Summing the harmonic series, using $\sum_{j\le m}j^{-1}\le1+\log m$, proves a block bound $O(N+q\log(2q))$. This is the required <separated reciprocal-distance sum> estimate, with its proof supplied here. There are $O(N/q+1)$ blocks, so
$$
|S|^2\ll N+\left(\frac Nq+1\right)(N+q\log(2q))
\ll(N^2/q+N+q)\log(2q).
$$
For $q<16$, the trivial bound $|S|\le N$ already implies this estimate after increasing the absolute constant. This completes the proof, including the small denominators and the possible common factor of $2$ and $q$.
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