Example 1 Find the largest positive integer $A$, such that in any permutation of $1,2, \cdots, 100$, there are 10 consecutive terms whose sum is not less than $A$. (22nd Polish Mathematical Olympiad Problem)
Let $a_{1}, a_{2}, \cdots, a_{100}$ be any permutation of $1,2, \cdots, 100$, and let $A_{i}=a_{i}+a_{i+1}+\cdots+a_{i+9}(i=1,2, \cdots, 91)$. Then $A_{1}=a_{1}+a_{2}+\cdots+a_{10}, A_{11}=a_{11}+a_{12}+\cdots+a_{20}, \cdots, A_{91}=a_{91}+a_{92}+\cdots+a_{100}$.

Notice that $A_{1}+A_{11}+\cdots+A_{91}=a_{1}+a_{2}+\cdots+a_{100}=5050$, so the average value of $A_{1}, A_{11}, \cdots, A_{91}$ is 505. Therefore, there exists at least one $i(1 \leqslant i \leqslant 91)$ such that $A_{i} \geqslant 505$.

When $A \geqslant 506$, consider the permutation: $(100,1,99,2,98,3,97,4, \cdots, 51,50)$, where each term satisfies: $a_{2 i-1}=101-i(i=1,2, \cdots, 50) ; a_{2 i}=i(i=1,2, \cdots, 50)$. Thus, $a_{2 i-1}+a_{2 i}=101, a_{2 i}+a_{2 i+1}=100$. At this point, it can be proven that for any $i(1 \leqslant i \leqslant 91)$, $A_{i} \leqslant 505<A$.

Actually, when $i$ is even, let $i=2 k$, then
$$\begin{aligned}
A_{2 k} & =a_{2 k}+a_{2 k+1}+\cdots+a_{2 k+9} \\
& =\left(a_{2 k}+a_{2 k+1}\right)+\left(a_{2 k+2}+a_{2 k+3}\right)+\cdots+\left(a_{2 k+8}+a_{2 k+9}\right) \\
& =100 \times 5=500<A
\end{aligned}$$

When $i$ is odd, let $i=2 k-1$, then
$$\begin{aligned}
A_{2 k-1} & =a_{2 k-1}+a_{2 k}+\cdots+a_{2 k+8} \\
& =\left(a_{2 k-1}+a_{2 k}\right)+\left(a_{2 k+1}+a_{2 k+2}\right)+\cdots+\left(a_{2 k+7}+a_{2 k+8}\right) \\
& =101 \times 5=505<A
\end{aligned}$$

Hence, $A_{\max }=505$.