The equality $(*)\lceil \frac{n^2}{1681}\rceil= \lfloor \frac{(n+1)^2}{1682}\rfloor$ is false for most natural numbers $n$ (in fact we have $\frac{n^2}{1681} \gt \frac{(n+1)^2}{1682}$ for $n\geq 3363$).
However (*) is true for the 272 successive integers in the interval $I=[1546,1816]$ except for $n=1681$. Is there a simple explanation to that fact ?
I’ve looked at the map $f(n)=\lceil \frac{n^2}{1681}\rceil= \lfloor \frac{(n+1)^2}{1682}\rfloor$ for $n\in I\setminus \lbrace 1681\rbrace$. The expression $f(n+1)-f(n)$ is nearly always equal to $2$, except in a few erratically placed spots where it is $1$ or $3$.
Maybe $1681=41^2$ plays a role here.
The difference (without floor and ceiling) between the two terms is $\frac{n^2-3362n-1681}{2827442}$. The apex of this parabola is at $1681$ and is $-1$ there and under "optimal conditions" for the floor and ceiling to wipe it away not only there, but also at the flat neighbourhood of it.
In fact notictíng that $$\lceil\frac {n^2}{1681}\rceil-2n+1681=\lceil\frac{(n-1681)^2}{1681}\rceil$$ and $$\lfloor \frac {(n+1)^2}{1682}\rfloor-2n+1681 =\lfloor \frac{(n-1681)^2}{1682}\rfloor +1$$ may make the existence of such a streak more easily visible