Example 4 Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for all real numbers $x_{1}, x_{2}, \cdots, x_{2000}$, we have
$$
\begin{array}{l}
\sum_{i=1}^{2005} f\left(x_{i}+x_{i+1}\right)+f\left(\sum_{i=1}^{2006} x_{i}\right) \\
\leqslant \sum_{i=1}^{2006} f\left(2 x_{i}\right) .
\end{array}
$$
Let $x_{1}=x_{2}=x$,
$$
x_{3}=x_{4}=\cdots=x_{2006}=0 \text {. }
$$

Substitute into equation (1) to get
$$
\begin{array}{l}
f(2 x)+f(x)+2003 f(0)+f(2 x) \\
\leqslant 2 f(2 x)+2004 f(0) .
\end{array}
$$

Thus, $f(x) \leqslant f(0)$.
Now let $x_{1}=x_{3}=\cdots=x_{2005}=\frac{x}{2}$,
$$
x_{2}=x_{4}=\cdots=x_{2006}=-\frac{x}{2} \text {. }
$$

Substitute into equation (1) to get
$$
\begin{array}{l}
2005 f(0)+f(0) \\
\leqslant 1003 f(x)+1003 f(-x),
\end{array}
$$

which means $2 f(0) \leqslant f(x)+f(-x) \leqslant f(x)+f(0)$. Therefore, $f(x) \geqslant f(0)$.
Hence, $f(x)=f(0)=c$ (constant).