Example 14 Try to find the maximum value of the following expression:
$$
P=|\cdots|\left|x_{1}-x_{2}\right|-x_{3}|-\cdots|-x_{1995} \mid,
$$

where $x_{1}, x_{2}, \cdots, x_{1995}$ are different natural numbers from 1 to 1995.
Solution: It is evident that when $x_{1} \geqslant 0, x_{2} \geqslant 0$, $\left|x_{1}-x_{2}\right|$ does not exceed the largest of $x_{1}$ and $x_{2}$; for $x_{1} \geqslant 0, x_{2} \geqslant$ $0, x_{3} \geqslant 0$, $\left|\left|x_{1}-x_{2}\right|-x_{3}\right|$ does not exceed the largest of $x_{1}$, $x_{2}$, and $x_{3}$. Therefore, $P \leqslant 1995$.

On the other hand, $P$ has the same parity as $x_{1}+x_{2}+\cdots+x_{1995}$, and $x_{1}+x_{2}+\cdots+x_{1995}=1+2+3+\cdots+1995$ is an even number, so $P \leqslant 1994$. In fact, $\left|\left|(4 k+1)-(4 k+3)\right|-(4 k+4)\right|-(4 k+2)=0$ holds for $k=0,1,2, \cdots$. Therefore, $1 \sim 1992$ can be processed according to the above method to yield a result of 0. And $\left|\left|1993-1994\right|-1995\right|=1994$. Hence, the maximum value of $P$ is 1994.