Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?

保留源文本的换行和格式，翻译结果如下：

Example 7. (IMO-19-2) In a sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?
Given that a junior high school student is not clear about the concept of "sequence", we can simply change the term "sequence" to "numbers arranged according to a certain rule".

Let the sequence be $a_{1}, a_{2}, \cdots a_{2}, \cdots, a_{0}$, i.e., $n \geq 17$. According to the problem:
$$
\begin{array}{l}
a_{k}+a_{k+1}+\cdots a_{k+0}0 . \quad(k \in N)
\end{array}
$$
(2) - (1) gives:
$$
a_{k+}+a_{k+1}+a_{k+0}+a_{k+10}>0 \text {. }
$$

From (3), we know that starting from the 8th term, the sum of any four consecutive terms is positive.
Then
$$
\begin{array}{l}
a_{8}+a_{8}+a_{10}+a_{11}>0, \\
a_{11}+a_{12}+a_{18}+a_{14}>0 .
\end{array}
$$
(4) + (5) gives: $a_{0}+a_{0}+a_{10}+2 a_{11}+a_{12}$
$$
+a_{13}+a_{14}>0 \text {. }
$$

Furthermore, from (1), we get: $a_{8}+a_{4}+a_{10}+a_{11}+a_{12}$
$$
+a_{13}+a_{14}0$.

Similarly, we get: $a_{12}>0, a_{18}>0$,
$$
a_{11}+a_{12}+a_{1 s}>0 \text {. }
$$

When $n \geqslant 17$, from (1) we get
$$
\begin{array}{l}
a_{11}+a_{13}+a_{18}+a_{14}+a_{17}+a_{10} \\
+a_{17}3$ and $a \in N$, sequences of the form:
$a, a,-\frac{5 a+1}{2}, a, a, a,-\frac{5 a+1}{2}$, $a, a,-\frac{5 a+1}{2}, a, a, a,-\frac{5 a+1}{2}, a, a$ are all sequences that satisfy the condition $n=16$.