Example 11 Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers. And $x_{1}$
$$
<x_{2}<\cdots<x_{6}<x_{7} \text {, and } x_{1}+x_{2}+\cdots+x_{7}=
$$
159. Find the maximum value of $x_{1}+x_{2}+x_{3}$.
(1997, Anhui Province Junior High School Mathematics Competition)
Solution: $\because 159=x_{1}+x_{2}+\cdots+x_{7}$
$$
\begin{array}{l}
\geqslant x_{1}+\left(x_{1}+1\right)+\left(x_{1}+2\right)+\cdots+\left(x_{1}+6\right) \\
=7 x_{1}+21, \\
\therefore x_{1} \leqslant 19 \frac{5}{7} .
\end{array}
$$

Therefore, the maximum value of $x_{1}$ is 19.
$$
\begin{array}{l}
\text { Also, } \because 19+x_{2}+x_{3}+\cdots+x_{7}=159, \\
\therefore 140 \geqslant x_{2}+\left(x_{2}+1\right)+\cdots+\left(x_{2}+5\right) \\
=6 x_{2}+15. \\
\therefore x_{2} \leqslant 20 \frac{5}{6}. \\
\end{array}
$$

Therefore, the maximum value of $x_{2}$ is 20.
$$
\begin{array}{l}
\text { From } 20+x_{3}+\cdots+x_{7}=140, \text { we have } \\
\begin{aligned}
120 & \geqslant x_{3}+\left(x_{3}+1\right)+\cdots+\left(x_{3}+4\right) \\
& =5 x_{3}+10 .
\end{aligned}
\end{array}
$$

Then $x_{3} \leqslant 22$,
Therefore, the maximum value of $x_{3}$ is 22.
Thus, the maximum value of $x_{1}+x_{2}+x_{3}$ is
$$
19+20+22=61.
$$