= Solution
Put $Z=\lfloor N^{2/5}\rfloor$ and $S=\sum_{N<n\le N+M}n^{-it}$. For any positive integer shift $h\le Z^2$, translating the interval changes its sum by at most $2h$. Averaging the shifts $h=xy$ gives the <bilinear shift averaging for a logarithmic phase> identity
$$
S=\frac1{Z^2}\sum_{N<n\le N+M}\sum_{x,y\le Z}(n+xy)^{-it}+O(Z^2).
$$
Since $xy/n\le N^{-1/5}$, the alternating <Taylor expansion> of the logarithm has remainder at most $(xy/n)^{r+1}/(r+1)$. Thus
$$
-t\log(1+xy/n)=\sum_{j=1}^r\frac{(-1)^jt}{jn^j}x^jy^j+O\bigl(tN^{-(r+1)/5}\bigr).
$$
For $r=\lfloor5.01\log t/\log N\rfloor$, we have $r+1>5.01\log t/\log N$, hence the error is at most $t^{-1/500}$. The <exponential function> on an imaginary argument changes by at most the change in that argument. Therefore
$$
\sum_{x,y\le Z}(n+xy)^{-it}=n^{-it}U(n)+O(Z^2t^{-1/500}).
$$
Use $Z^2\asymp N^{4/5}$ and $N<n\le N+M\le2N$. Taking absolute values proves
$$
\boxed{|S|\ll M\max_{N\le n\le2N}\frac{|U(n)|}{N^{4/5}}+N^{4/5}+Mt^{-1/500}.}
$$
The boundary error comes from integer shifts, so this argument also covers intervals shorter than a shift. Here $x,y$ range over positive integers; no zero term is needed.
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