China Hong Kong Mathematical Olympiad

The answer is $\frac{p-1}{4}$.
Note that $1^2, 2^2, \dots, \left(\frac{p-1}{2}\right)^2$ are pairwise incongruent modulo $p$ since
$$x^2 \equiv y^2 \pmod{p} \Rightarrow x \equiv \pm y \pmod{p}.$$
As there are exactly $\frac{p-1}{2}$ nonzero quadratic residues, $1^2, 2^2, \dots, \left(\frac{p-1}{2}\right)^2$ are all the nonzero quadratic residues modulo $p$.
Since $p \equiv 1 \pmod 4$, $-1$ is a quadratic residue modulo $p$. Therefore, $r$ is a quadratic residue modulo $p$ if and only if $-r$ is a quadratic residue modulo $p$. This shows the nonzero quadratic residues come in pairs such that the numbers in each pair are negative of each other modulo $p$. Thus, $\left\{ \frac{k^2}{p} \right\}$ for $k = 1, 2, \dots, \frac{p-1}{2}$ can be partitioned into $\frac{p-1}{4}$ pairs, such that the sum of each pair is $\frac{r + (p-r)}{p} = 1$. Thus, the sum is equal to $\frac{p-1}{4}$.