Problem 1320

1. We start with the given function $F(p) = \sum_{k=1}^{\frac{p-1}{2}} k^{120}$. We need to evaluate this sum modulo $p$.

2. Consider the sum $2F(p) = \sum_{k=1}^{\frac{p-1}{2}} (k^{120} + (p-k)^{120})$. Since $(p-k) \equiv -k \pmod{p}$, we have:
$$(p-k)^{120} \equiv (-k)^{120} \equiv k^{120} \pmod{p}$$
Therefore,
$$2F(p) \equiv \sum_{k=1}^{\frac{p-1}{2}} (k^{120} + k^{120}) \equiv \sum_{k=1}^{\frac{p-1}{2}} 2k^{120} \equiv 2 \sum_{k=1}^{\frac{p-1}{2}} k^{120} \pmod{p}$$
This simplifies to:
$$2F(p) \equiv \sum_{k=1}^{p-1} k^{120} \pmod{p}$$

3. Let $r$ be a primitive root modulo $p$. The sequence $r, r^2, \ldots, r^{p-1}$ is a permutation of $1, 2, \ldots, p-1$ modulo $p$. Thus,
$$\sum_{k=1}^{p-1} k^{120} \equiv \sum_{i=1}^{p-1} r^{120i} \pmod{p}$$

4. We can use the formula for the sum of a geometric series. Since $r^{120(p-1)} \equiv 1 \pmod{p}$ (because $r^{p-1} \equiv 1 \pmod{p}$), we have:
$$\sum_{i=1}^{p-1} r^{120i} = r^{120} \frac{r^{120(p-1)} - 1}{r^{120} - 1} \equiv 0 \pmod{p} \quad \text{if } r^{120} \not\equiv 1 \pmod{p}$$
This implies:
$$\sum_{i=1}^{p-1} r^{120i} \equiv 0 \pmod{p}$$

5. Therefore, if $p-1 \nmid 120$, then $r^{120} \not\equiv 1 \pmod{p}$ and:
$$F(p) \equiv 0 \pmod{p}$$
Hence,
$$f(p) = \frac{1}{2} - \left\{ \frac{F(p)}{p} \right\} = \frac{1}{2} - 0 = \frac{1}{2}$$

6. If $p-1 \mid 120$, then $r^{120} \equiv 1 \pmod{p}$. In this case:
$$\sum_{i=1}^{p-1} r^{120i} = \sum_{i=1}^{p-1} 1 = p-1$$
Therefore,
$$F(p) = \frac{1}{2} \sum_{k=1}^{p-1} k^{120} \equiv \frac{1}{2} (p-1) \pmod{p}$$
Hence,
$$f(p) = \frac{1}{2} - \left\{ \frac{F(p)}{p} \right\} = \frac{1}{2} - \frac{p-1}{2p} = \frac{1}{2} - \frac{1}{2} + \frac{1}{2p} = \frac{1}{2p}$$

The final answer is $\boxed{\frac{1}{2}}$ if $p-1 \nmid 120$ and $\boxed{\frac{1}{2p}}$ if $p-1 \mid 120$.