Example 3-30 Let $a_{1}, a_{2}, \cdots, a_{100}$ be a sequence composed of 1s and 2s, and it is known that the sum of any 10 consecutive numbers starting from any number does not exceed 16. That is, for $1 \leqslant i \leqslant 91$, we always have
$$
a_{i}+a_{i+1}+\cdots+a_{i+9} \leqslant 16
$$

Then there exist at least $h$ and $k, k>h$, such that
$$
a_{h}+a_{h+1}+\cdots+a_{k}=39
$$
The sequence $s_{1}=a_{1}, s_{2}=a_{1}+a_{2}, \cdots, s_{100}=a_{1}+a_{2}+\cdots+a_{100}$. Since each $a_{i}$ is a positive integer, we have
$$
s_{1}<s_{2}<\cdots<s_{100}
$$

Moreover,
$$
\begin{aligned}
s_{100}= & \left(a_{1}+a_{2}+\cdots+a_{10}\right)+\left(a_{11}+a_{12}+\cdots+a_{20}\right) \\
& +\cdots+\left(a_{91}+a_{92}+\cdots+a_{100}\right)
\end{aligned}
$$

Thus, according to the assumption,
$$
s_{100} \leqslant 10 \times 16=160
$$

Consider the sequence
$$
\underbrace{s_{1}, s_{2}, \cdots, s_{100}, s_{1}+39, s_{2}+39, \cdots, s_{100}+39}
$$

which has 200 terms. The last term $s_{100}+39 \leqslant 160+39=199$. However, the sequence has 200 terms, which are integers from 1 to 199. By the pigeonhole principle, there must be two terms that are equal. But the first 100 terms are monotonically increasing, and the last 100 terms are also monotonically increasing, so there exist $h$ and $k$ such that
$$
s_{h}=s_{k}+39, \quad 1 \leqslant h, k \leqslant 100
$$

Thus,
$$
s_{h}-s_{k}=39
$$

which implies
$$
a_{1}+a_{2}+\cdots+a_{k}-\left(a_{1}+a_{2}+\cdots+a_{k}\right)=39
$$

or
$$
a_{k+1}+a_{k+2}+\cdots+a_{h}=39
$$