$8 \cdot 18$ In a finite 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?
[Solution] Let $a_{1}, a_{2}, \cdots, a_{n}$ be the sequence satisfying the given conditions. If $70$ is known,
$$
a_{k}+a_{k+1}+a_{k+2}+a_{k+3}>0 .
$$

If $40$, and by $a_{k}+a_{k+1}+\cdots+a_{k+6}0, a_{8}+a_{9}+a_{10}0$.
Similarly, $a_{12}>0, a_{13}>0$, but from (2) we get $a_{11}+a_{12}+a_{13}<0$, which is a contradiction! An example for $n=16$:
$$
5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5 \text {. }
$$

In conclusion, the sequence satisfying the given conditions can have at most 16 terms.