= Solution
Consider the graph
$$
\Gamma=\{(x,f(x)):1\leq x\leq N\}\subseteq\mathbb Z^2.
$$
The hypothesis says exactly that $\Gamma$ has at least $\delta N^3$ <additive quadruples>. By the <Balog-Szemerédi-Gowers theorem>, it contains $\Gamma'$ with $|\Gamma'|\gg_\delta N$ and $|\Gamma'+\Gamma'|\ll_\delta|\Gamma'|$.
Part b gives a <Freiman s-isomorphism> from $\Gamma'$ to a set $D\subseteq\mathbb Z$, where it is enough to take any fixed $s\geq2$. The set $D$ has bounded doubling, with a bound depending only on $\delta$. Part c therefore provides a proper <generalized arithmetic progression> $Q$ of rank $O_\delta(1)$ such that
$$
|D\cap Q|\gg_\delta|D|\gg_\delta|Q|.
$$
Write $Q$ as a proper parameter box. Since $|Q|\gg_\delta N$, one of its $O_\delta(1)$ side lengths tends to infinity with $N$. Averaging over all lines parallel to that side gives a line on which $D\cap Q$ has density bounded below in terms of $\delta$. The <Szemerédi theorem> quoted in the question then gives, once $N$ is sufficiently large in terms of $k$ and $\delta$, a nonconstant $k$-term <arithmetic progression> in $D$.
The inverse Freiman isomorphism sends it to a $k$-term arithmetic progression in $\Gamma'\subseteq\Gamma$, because each relation between three consecutive terms is an additive-quadruple relation. Write this progression as
$$
(x_0,y_0),\ (x_0+d,y_0+e),\ldots,
(x_0+(k-1)d,y_0+(k-1)e).
$$
Its first-coordinate difference cannot be zero: the graph of a function has only one point above each $x$. Hence $d\ne0$. On the nonconstant progression
$$
P=\{x_0,x_0+d,\ldots,x_0+(k-1)d\}\subseteq\{1,\ldots,N\},
$$
we have
$$
f(x)=\frac edx+\left(y_0-\frac edx_0\right).
$$
Taking $a=e/d$ and $b=y_0-(e/d)x_0$, both in $\mathbb Q$, proves the claim.
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