Lemma: For $a,b$ real numbers holds $\{a-b\}+\{b-a\}\in \{0,1\}$, therefore $\{a-b\}+\{b-a\} \leq 1$ Proof: Trivial. Assume that for $y=x_i, 1\leq i \leq n-1$ we have $\sum_{i=1}^{n} \{y-x_i \}>\frac{n-1}{2}$ Adding these sums we have that $\sum_{1\leq i<j \leq n-1} \left(\{x_i-x_j\}+\{x_j-x_i\}\right)+\sum_{i=1}^{n-1} \{x_i-x_n \}>\frac{(n-1)^2}{2}$ Using the lemma we have $\frac{(n-1)^2}{2}\leq n-1-\sum_{i=1}^{n-1} \{x_n -x_i \}+\dbinom{n-1}{2} \Longleftrightarrow \sum_{i=1}^{n-1} \{x_n -x_i \}\leq \frac{n-1}{2}.$ $\blacksquare$

I am not sure if this is correct but it is very similar to @XbenX's solution. Assume that $\forall i = 1, 2, …, n$ $\sum_{j=1}^n \{x_i-x_j\} > \frac{n-1}{2}$. Adding these sums we have that $\sum_{1\leq i<j \leq n} \left(\{x_i-x_j\}+\{x_j-x_i\}\right) > \frac{n(n-1)}{2}$ . Given lemma says $ \frac{n(n-1)}{2}=\dbinom{n}{2} \ge LHS > \frac{n(n-1)}{2}$. Contradiction.