AMC 12 Mock Exam 5 Problem 19

Let $f(k)=\left\lfloor\frac{k^2}{2026}\right\rfloor$. Since $(k+1)^2-k^2=2k+1$:

For $1\le k\le1012$, $2k+1\le2025<2026$, so $f(k+1)-f(k)\in\{0,1\}$. Hence $f(1),\dots,f(1013)$ take every integer value from $f(1)=0$ to $f(1013)=\left\lfloor\frac{1026169}{2026}\right\rfloor=506$: that is $507$ values.

For $k\ge1013$, $2k+1\ge2027>2026$, so $f(k+1)>f(k)$. Hence $f(1014),\dots,f(2026)$ are $1013$ distinct values, all larger than $f(1013)$.

The total is $507+1013=\boxed{1520}$.