Example 5. Find the largest integer $A$, such that for any permutation of all natural numbers from 1 to 100, there are 10 consecutive numbers whose sum is greater than or equal to $A$.
[22nd Polish Mathematical Olympiad]
This problem, although it is to find the maximum integer $A$, is actually an existence problem, i.e., there exists a maximum integer $A$ such that the sum of 10 consecutive numbers is not less than $A$.

Solution: Let $T=\left(a_{1}, a_{2}, \cdots, a_{100}\right)$ be a permutation of the natural numbers from 1 to 100.
Consider the sum of 10 consecutive terms
$$
\sum_{k=1}^{10} a_{a+k}, \quad n=1,2, \cdots, 90 .
$$

This is a finite set, and in a finite set, there must be a maximum number, denoted as
$$
A_{\mathrm{T}}=\max _{1 \leqslant n \leqslant 90}^{\text {io }} \sum_{\mathrm{D}=1} a_{n}+x_{0}
$$
Any other sum of 10 consecutive terms does not exceed $A_{x}$.
By the definition of $A_{\Gamma}$, we have
$$
\begin{array}{l}
A_{\mathrm{T}} \geqslant a_{1}+a_{2}+\cdots+a_{10} \\
A_{\mathrm{T}} \geqslant a_{11}+a_{12}+\cdots+a_{20}, \\
\quad \cdots \cdots \\
A_{\mathrm{X}} \geqslant a_{61}+a_{82}+\cdots+a_{100}
\end{array}
$$

Adding these up, we get
$$
10 A_{\mathrm{r}} \geqslant \sum_{1=1}^{100} a_{1}=5050,
$$
$\rightarrow \operatorname{mosers}$
$$
A_{\mathrm{T}}>505 \text {. }
$$

According to the problem, we need to find the smallest $A_{x}$ among all permutations $T$, i.e.,
$$
A=\min A_{\mathrm{r}}
$$

Below, we can find a permutation $T^{\prime}$ such that for the permutation $T^{\prime}$, its $A_{\mathrm{T}}{ }^{\prime} \leqslant 505$.
For example, we can arrange the numbers from 1 to 100 as follows:
$$
\begin{array}{c}
T^{\prime}=(100,1,99,2,98,3,97, \\
4, \cdots, 51,50),
\end{array}
$$

which satisfies
$$
\begin{array}{l}
a_{2 \Omega+1}=100-n, 0 \leqslant n \leqslant 49, \\
a_{2 \mathrm{n}}=n, 1 \leqslant n \leqslant 50 .
\end{array}
$$

At this point, we have
$$
\begin{array}{l} 
a_{2 k}+a_{2 k+1}+\cdots+a_{2 k+2} \\
=\left(a_{2 k}+a_{2 k+2}+a_{2 k+4}+a_{2 k+0}\right. \\
\left.+a_{2 k+8}\right)+\left(a_{2 k+1}+a_{2 k+8}+a_{2 k+5}\right. \\
\left.+a_{2 k+1}+a_{2 k+0}\right) \\
=(k+k+1+k+2+k+3+k+4) \\
+[100-k+100-(k+1)+100-(k \\
+2)+100-(k+3)+100-(k+4)] \\
= 500 \\
a_{2 k+1}+a_{2 k+2}+\cdots+a_{2 k+10} \\
=\left(a_{2 k}+a_{2 k+1}\right)+\left(a_{2 k+2}+a_{2 k+8}\right)+\cdots
\end{array}
$$
$$
\begin{aligned}
& +\left(a_{2 k+8}+a_{2 k+2}\right)+a_{2 k+10}-a_{2 k} \\
= & 500+k+5-5 \\
= & 505 .
\end{aligned}
$$

Thus, $A \mathrm{x}^{\prime} \leqslant 505$.
From (1) and (2), we get $A=505$.