$15 \cdot 34$ Find the largest integer $A$, such that for any permutation of all natural numbers from 1 to 100, there are 10 consecutive positions whose sum is greater than or equal to $A$.
(Polish Mathematical Competition, 1970)
[Solution] Let $\sigma=\left(a_{1}, a_{2}, \cdots, a_{100}\right)$ be any permutation of all natural numbers from 1 to 100, and let
$$
A_{\sigma}=\max _{0 \leqslant n \leqslant 90} \sum_{\mathrm{k}=1}^{10} a_{n+k} .
$$

Thus, in the permutation $\sigma$, there are 10 consecutive terms whose sum equals $A_{\sigma}$, and the sum of any other 10 consecutive terms does not exceed $A_{\sigma}$. Therefore, the problem reduces to finding the number
$$
A=\min _{\sigma} A_{\sigma} .
$$

By the definition of $A_{\sigma}$, we have
$$
\begin{array}{l}
A_{\sigma} \geqslant a_{1}+a_{2}+\cdots+a_{10}, \\
A_{\sigma} \geqslant a_{11}+a_{12}+\cdots+a_{20}, \\
\cdots \cdots \\
A_{\sigma} \geqslant a_{91}+a_{92}+\cdots+a_{100} .
\end{array}
$$

Adding both sides of these inequalities, we get
$$
\begin{aligned}
10 A_{\sigma} & \geqslant a_{1}+a_{2}+\cdots+a_{100} \\
& =1+2+\cdots+100=5050 .
\end{aligned}
$$

Therefore, for any permutation $\sigma$, the inequality
$$
A_{\sigma} \geqslant 505
$$
holds.

From (2), we also have $A \geqslant 505$.
Now consider the following permutation $\tau=\left(a_{1}, a_{2}, \cdots, a_{100}\right)$ of the natural numbers from 1 to 100:
$100,1,99,2, \quad 98,3,97,4, \cdots, 51,50$.
This permutation can be described by the following relations:
$$
\begin{array}{ll}
a_{2 n+1}=100-n, & \text { (when } 0 \leqslant n \leqslant 49 \text {)}, \\
a_{2 n}=n, & \text { (when } 1 \leqslant n \leqslant 50 \text {). }
\end{array}
$$

We prove that the sum of any 10 consecutive terms in this permutation does not exceed 505.
Indeed, if the first term of these 10 consecutive terms has an even index $2 k$, then
$$
\begin{aligned}
S_{1}= & a_{2 k}+a_{2 k+1}+\cdots+a_{2 k+9} \\
= & \left(a_{2 k}+a_{2 k+2}+\cdots+a_{2 k+8}\right)+\left(a_{2 k+1}+a_{2 k+3}+\cdots+a_{2 k+9}\right) \\
= & {[k+(k+1)+\cdots+(k+4)]+[(100-k)+(100} \\
& -k-1)+\cdots+(100-k-4)] \\
= & 500 .
\end{aligned}
$$

If the first term of these 10 consecutive terms has an odd index $2 k+1$, then
$$
\begin{aligned}
S_{2} & =a_{2 k+1}+a_{2 k+2}+\cdots+a_{2 k+10} \\
& =\left(a_{2 k}+a_{2 k+1}+\cdots+a_{2 k+9}\right)+a_{2 k+10}-a_{2 k} \\
& =S_{1}+(k+5)-k \\
& =505 .
\end{aligned}
$$

Thus, we have proven that the sum of any 10 consecutive terms in the permutation $\tau$ does not exceed 505, and it can equal 505.
Therefore, from (2), we get $A \leqslant 505$.
Thus, from (3) and (4), we have $A=505$.