Solution: Since this inequality is cyclic rather than symmetric, we have specified the order of the variables. If we rely on the relation $\left(x_{i}-1\right)\left(x_{i}+1\right) \leq 0$, we will not succeed.

Intuitively, we believe that the expression reaches its maximum value when the sequence $x_{1}, x_{2}, \cdots, x_{2005}$ alternates between 1 and -1. In this case,
$$P=1 \cdot(-1)+(-1) \cdot 1+\cdots+(-1) \cdot 1+1 \cdot 1=-2003$$

A precise proof of this conjecture is not so obvious. Although the following solution is simple, it is indeed difficult if you do not have knowledge of convex functions.

First, we note that if $x \in[p, q]$, then each linear function $f(x)=a x+b$ or quadratic function $f(x)=a x^{2}+b x+c$ has the following important property:
$$\begin{array}{l} \max _{x \in[p, q]} f(x)=\max \{f(p), f(q)\} \\ \min _{x \in[p, q]} f(x)=\min \{f(p), f(q)\} \end{array}$$

Notice that $P=P\left(x_{1}\right)$ is a linear function of $x_{1}$, so according to the property of linear functions, $P$ reaches its minimum value if and only if $x_{1} \in\{-1,1\}$. Similarly, for other variables, we get that $P$ reaches its minimum value if and only if $x_{k} \in\{-1,1\} \quad(k=1,2, \cdots, 2005)$. In this case, we will prove that $P \geq-2003$. Indeed, there must be at least one $k(k \in \mathbb{N}, 1 \leq k \leq 2005)$ such that $x_{k} x_{k+1} \geq 0$. This means $x_{k} x_{k+1}=1$, so $\sum_{k=1}^{2005} x_{k} x_{k+1} \geq-2003$