Notice that,
$$\begin{array}{l} S=\sum_{k=1}^{n}\left((k-1) x_{k}-(n-k) x_{k}\right) \\ =\sum_{k=1}^{n}\left(x_{k}(2 k-n-1)\right) \\ \leqslant \sum_{k \geq \frac{n+1}{2}}\left(x_{k}(2 k-n-1)\right) . \end{array}$$

When $n$ is odd,
$$\begin{aligned} S & \leqslant \sum_{k=\frac{n+1}{2}}^{n}\left(x_{k}(2 k-n-1)\right) \\ & \leqslant \sum_{k=\frac{n+1}{2}}^{n}(2 k-n-1) \\ & =\frac{n^{2}-1}{4}, \end{aligned}$$

When $x_{1}=x_{2}=\cdots=x_{\frac{n-1}{2}}=0, x_{\frac{n+1}{2}}=x_{\frac{n+3}{2}}=\cdots=x_{n}=1$, the equality can be achieved;
When $n$ is even,
$$\begin{array}{l} S \leqslant \sum_{k=\frac{n+2}{2}}^{n}\left(x_{k}(2 k-n-1)\right) \\ \leqslant \sum_{k=\frac{n+2}{2}}^{n}(2 k-n-1)=\frac{n^{2}}{4}, \end{array}$$

When $x_{1}=x_{2}=\cdots=x_{\frac{n}{2}}=0, x_{\frac{n}{2}+1}=x_{\frac{n}{2}+2}=\cdots=x_{n}=1$, the equality can be achieved.

Thus, the maximum possible value is $\left[\frac{n^{2}}{4}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$.