= Small Type I sums for binary digit parity
{title2=$N^{-1}\sum_{m\le N^{1/100}}|\sum_{n\in I_m}(-1)^{s_2(mn)}|\ll N^{\sigma+1/100-1}$}
Let $f(r)=(-1)^{s_2(r)}$ be the <Thue–Morse sequence>, let $e(t)=e^{2\pi it}$, and let $N\geq2$. For each positive integer $m\leq N^{1/100}$, take an integer interval $I_m\subseteq[1,N/m]$. The range condition ensures that every product $mn$ remains at most $N$.
Choose the integer interval $J_m$ from $m\min I_m$ to $m\max I_m$, omitting an empty $I_m$. <Character orthogonality> gives
$$
\sum_{n\in I_m}f(mn)=\frac1m\sum_{a=0}^{m-1}\sum_{r\in J_m}f(r)e(ar/m).
$$
The <uniform interval cancellation for binary digit parity> bounds each inner <exponential sum> by $CN^\sigma$, with $\sigma=2-\tfrac34\log_2 3<1$. The average over $a$ does not increase that bound. Therefore
$$
N^{-1}\sum_{m\leq N^{1/100}}\left|\sum_{n\in I_m}f(mn)\right|\leq C N^{\sigma+1/100-1}.
$$
The exponent is negative, so this normalized <Type I sum> bound is smaller than every fixed negative power of $\log N$ for large $N$. No cancellation between the outer terms is needed; each inner progression is controlled separately.
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