Define \(x_0=0, x_{n+1}=1, y_k=x_{k+1}-x_k\) for \(k=0,1,2, \ldots, n\). Then \(y_0=y_n=0\). Assume the contrary, i.e. that \(\left|x_{j+1}+x_{j-1}-2 x_j\right|=\left|y_j-y_{j-1}\right|<\frac{4}{n^2}\) for each \(j \in\{1,2, \ldots, n\}\). Adding the inequalities \(y_j-y_{j-1}<\frac{4}{n^2}\) for \(j=1,2, \ldots, k\) we get \(y_k<\frac{4 k}{n^2}\). Similarly, by adding \(y_j-y_{j+1}<\frac{4}{n^2}\) for \(j=k, k+1, \ldots, n-1\) we get \(y_k<\frac{4(n-k)}{n^2}\). Consider the following two cases:

**Case 1**. If \(n\) odd then

\[

\begin{aligned}

1 &=y_0+y_1+\cdots+y_n=\left(y_1+\cdots+y_{\frac{n-1}{2}}\right)+\left(y_{\frac{n+1}{2}}+\cdots+y_n\right) \\

&<\frac{4}{n^2} \cdot \frac{1}{2} \cdot\left(\frac{n-1}{2} \cdot \frac{n+1}{2} \cdot 2\right)=\frac{n^2-1}{n^2}

\end{aligned}

\]

a contradiction.

**Case 2. **If \(n\) is even, we get a contradiction as well since

\[

\begin{aligned}

1 &=y_0+y_1+\cdots+y_n=\left(y_1+\cdots+y_{\frac{n}{2}-1}\right)+\left(y_{\frac{n}{2}}+\cdots+y_{n-1}\right) \\

&<\frac{4}{n^2} \cdot \frac{1}{2} \cdot\left(\left(\frac{n}{2}-1\right) \cdot \frac{n}{2}+\frac{n}{2} \cdot\left(\frac{n}{2}+1\right)\right)=1

\end{aligned}

\]