One, (40 points) Given $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers. Try to find
$$
E\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}+\sum_{i=1}^{n} x_{i}
$$

the minimum value.
Notice,
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1} \\
=\frac{1}{2}\left[x_{1}^{2}+\sum_{i=1}^{n-1}\left(x_{i}+x_{i+1}\right)^{2}+x_{n}^{2}\right] .
\end{array}
$$

Discuss in two cases.
(1) When $n=2 k$, by
$$
\begin{array}{l}
{\left[1+\left(\frac{k+1}{k}\right)^{2}+1+\cdots+\left(\frac{k+1}{k}\right)^{2}+1\right] \cdot} \\
{\left[x_{1}^{2}+\sum_{i=1}^{2 k-1}\left(x_{i}+x_{i+1}\right)^{2}+x_{2 k}^{2}\right]} \\
\geqslant\left(\frac{2 k+1}{k}\right)^{2}\left(\sum_{i=1}^{2 k} x_{i}\right)^{2},
\end{array}
$$

we know
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1} \geqslant \frac{2 k+1}{2 k(k+1)}\left(\sum_{i=1}^{2 k} x_{i}\right)^{2} \\
\Rightarrow E\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
\quad \geqslant \frac{2 k+1}{2 k(k+1)}\left(\sum_{i=1}^{2 k} x_{i}\right)^{2}+\sum_{i=1}^{2 k} x_{i} \\
\geqslant-\frac{k(k+1)}{2(2 k+1)}=\frac{n(n+2)}{8(n+1)} .
\end{array}
$$

Equality holds when $\sum_{i=1}^{2 k} x_{i}=-\frac{k(k+1)}{2 k+1}$, thus,
$$
\begin{array}{l}
x_{2 i-1}=-\frac{k+1-i}{2 k+1}(1 \leqslant i \leqslant k), \\
x_{2 i}=-\frac{i}{2 k+1}(1 \leqslant i \leqslant k) .
\end{array}
$$
(2) When $n=2 k+1$, by
$$
\begin{array}{l}
(1+1+\cdots+1)\left[x_{1}^{2}+\sum_{i=1}^{2 k}\left(x_{i}+x_{i+1}\right)^{2}+x_{2 k+1}^{2}\right] \\
\geqslant 4\left(\sum_{i=1}^{2 k+1} x_{i}\right)^{2}, \\
\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1} \geqslant \frac{1}{k+1}\left(\sum_{i=1}^{2 k+1} x_{i}\right)^{2} \\
\Rightarrow E\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
\quad \geqslant \frac{1}{k+1}\left(\sum_{i=1}^{2 k+1} x_{i}\right)^{2}+\sum_{i=1}^{2 k+1} x_{i} \\
\quad \geqslant-\frac{k+1}{4}=-\frac{n+2}{8} .
\end{array}
$$

we know
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1} \geqslant \frac{1}{k+1}\left(\sum_{i=1}^{2 k+1} x_{i}\right)^{2} \\
\Rightarrow E\left(x_{1}, x_{2}, \cdots, x_{n}\right) \\
\quad \geqslant \frac{1}{k+1}\left(\sum_{i=1}^{2 k+1} x_{i}\right)^{2}+\sum_{i=1}^{2 k+1} x_{i} \\
\quad \geqslant-\frac{k+1}{4}=-\frac{n+2}{8} .
\end{array}
$$

Equality holds when $\sum_{i=1}^{2 k+1} x_{i}=-\frac{k+1}{2}$, thus,
$$
\begin{array}{l}
x_{1}=x_{3}=\cdots=x_{2 k+1}=-\frac{1}{2}, \\
x_{2}=x_{4}=\cdots=x_{2 k}=0 .
\end{array}
$$