First of all, the maximum $\sum\limits_{i=1}^{n-1}\lfloor x_{i+1}-x_i \rfloor \\ \leq \sum\limits_{i=1}^{n-1}\ (x_{i+1}-x_i) \\ = x_n -x_1 $ Because $ \sum\limits_{i=1}^{n-1}\lfloor x_{i+1}-x_i \rfloor $ is an integer so, $ \sum\limits_{i=1}^{n-1}\lfloor x_{i+1}-x_i \rfloor\\ \leq \lfloor x_n-x_1 \rfloor\\ \leq n-1 $ The equal sign hold when $x_i=i$ The minimum: $\sum\limits_{i=1}^{n-1}\lfloor x_{i+1}-x_i \rfloor \\ > \sum\limits_{i=1}^{n-1}\ (x_{i+1}-x_i-1) \\ = x_n-x_1-n+1 $ Because $ \sum\limits_{i=1}^{n-1}\lfloor x_{i+1}-x_i \rfloor $ is an integer so, $ \sum\limits_{i=1}^{n-1}\lfloor x_{i+1}-x_i \rfloor\\ \geq \lceil x_n-x_1 \rceil-n+1\\ \geq 1-n-n+1\\ =-2n+2 $ equality hold when $x_{n+1-i}=i+i\epsilon$ i.e. $x_n =1+\epsilon\\ x_{n-1}=2+2\epsilon\\ x_{n-2}=3+3\epsilon\\ ...\\ x_1=n+n\epsilon $ Where $\epsilon$ is a sufficiently small positive real number

Of course there are (infinitely many) other cases when extrema are reached.