Example 13 (1993 National Training Team Selection Exam Question) For a prime $p \geqslant 3$, define
$$
F(p)=\sum_{k=1}^{\frac{p-1}{2}} k^{120}, f(p)=\frac{1}{2}-\left\{\frac{F(p)}{p}\right\},
$$

where $\{x\}=x-[x]$ denotes the fractional part of $x$. Find the value of $f(p)$.
Solve the division with remainder of 120 by $p-1$:
$120=q(p-1)+r, 0 \leqslant r \leqslant p-2$.
Since the prime $p \geqslant 3$, $p$ is odd. Because 120 and $p-1$ are even, $r$ is even.
Define $G(p)=\sum_{k=1}^{\frac{p-1}{2}} k^{r}$.
According to Fermat's Little Theorem, for $k=1,2, \cdots, \frac{p-1}{2}$, we have $k^{p-1} \equiv 1(\bmod p)$, so
$F(p) \equiv G(p)(\bmod p)$.
$f(p)=\frac{1}{2}-\left\{\frac{F(p)}{p}\right\}=\frac{1}{2}-\left\{\frac{G(p)}{p}\right\}$.
We will discuss the following two cases.
Case I: $r=0$.
In this case, $G(p)=\sum_{k=1}^{\frac{p-1}{2}} k^{0}=\frac{p-1}{2}$.
$$
f(p)=\frac{1}{2}-\left\{\frac{G(p)}{p}\right\}=\frac{1}{2}-\frac{p-1}{2 p}=\frac{1}{2 p} \text {. }
$$

Case II: $r \neq 0$.
Since $r$ is even, for $\bmod p$, we have
$G(p)=1^{r}+2^{r}+\cdots+\left(\frac{p-1}{2}\right)^{r}$
-
$$
\begin{aligned}
& \equiv(p-1)^{r}+(p-2)^{r}+\cdots+\left(p-\frac{p-1}{2}\right)^{r} \\
& =(p-1)^{r}+(p-2)^{r}+\cdots+\left(\frac{p+1}{2}\right)^{r}(\bmod p) . \\
2 G(p) & =G(p)+G(p) \equiv \sum_{i=1}^{p-1} i^{r}(\bmod p) .
\end{aligned}
$$

Since the congruence equation $x^{r} \equiv 1(\bmod p)$ has at most $r$ distinct solutions, and $0 \leqslant r \leqslant p-2$, there exists at least one $a \in\{1,2, \cdots, p-1\}$ such that $a^{r} \neq 1(\bmod p)$.
We have
$$
2 a^{r} G(p)=(1 \cdot a)^{r}+(2 \cdot a)^{r}+\cdots+((p-1) a)^{r} \equiv 2 G(p)(\bmod p) \text {. }
$$
(This is because $1 \cdot a, 2 \cdot a, \cdots,(p-1) a$ are pairwise incongruent modulo $p$, forming a complete residue system modulo $p$.)
Thus, $2\left(a^{r}-1\right) G(p) \equiv 0(\bmod p)$.
Since $2\left(a^{r}-1\right) \neq 0(\bmod p)$, we have
$G(p) \equiv 0(\bmod p)$.
In this case, $f(p)=\frac{1}{2}-\left\{\frac{G(p)}{p}\right\}=\frac{1}{2}$.
Now, we seek primes $p \geqslant 3$ that belong to Case I, i.e., primes $p \geqslant 3$ such that $p-1 \mid 120$. These primes are $3,5,7,11,13,31,41,61$.

All other primes belong to Case II.
In summary, the answer to this problem is