For any permutation $\sigma$ of $\{ 1, \dots ,n \}$ there exists $k$ such that $|k- \sigma (k)| \le \frac{n}{2}$ Let $n \ge 2$ be an even number.
Is the following true?

For any permutation $\sigma$ of $\{ 1, \dots ,n \}$ there exists $k$ such that $$|k- \sigma (k)| \le \frac{n}{2}$$

and is $n/2$ optimal?
I mean, is it true that
$$\max_{\sigma \in S_n} ( \min_{1 \le k \le n} |k- \sigma (k)| ) = \frac{n}{2}$$
Considering the permutation
$$\sigma^*_n(k) = \begin{cases}k + \frac{n}{2} & \ \mathrm{if} \ k \le \frac{n}{2} \\
k - \frac{n}{2} & \ \mathrm{if} \ k > \frac{n}{2}\end{cases}$$
every element of $\{ 1, \dots ,n\}$ is mapped to something disting at least $n/2$. This shows that
$$\max_{\sigma \in S_n} ( \min_{1 \le k \le n} |k- \sigma (k)| ) \ge
\min_{1 \le k \le n} |k- \sigma^*_n (k)| =\frac{n}{2}$$
Can we do better or is this the best case? 

It is optimal and we can take $k = \frac{n}{2}$: as $1 \leqslant \sigma(k) \leqslant n$, we have $1 - \frac{n}{2} \leqslant \sigma(\frac{n}{2}) - \frac{n}{2} \leqslant \frac{n}{2}$, thus $|\frac{n}{2} - \sigma(\frac{n}{2})| \leqslant \frac{n}{2}$.
