$11 \cdot 22$ Find 8 positive integers $n_{1}, n_{2}, \cdots, n_{8}$, such that they have the following property: For each integer $k, -1985 \leqslant k \leqslant 1985$, there are 8 integers $a_{1}, a_{2}, \cdots, a_{8}$, each $a_{i}$ belonging to the set $\{-1,0,1\}$, such that $k=\sum_{i=1}^{8} a_{i} n_{i}$.
(26th International Mathematical Olympiad Candidate Problem, 1985)
［Solution］Consider the ternary representation of 1985
$$
1985=2 \cdot 3^{6}+2 \cdot 3^{5}+3^{3}+3^{2}+3+2 \text {. }
$$

However, by the problem's condition, the coefficients in the representation cannot be 2, and since 2=3-1, then 1985 can be expressed as
$$
1985=3^{7}-3^{5}+3^{4}-3^{3}-3^{2}-3-1
$$

Similarly, we have $\quad 1984=3^{7}-3^{5}+3^{3}+3^{2}+3+1$.
Now starting from 1984, subtract 1 successively, two cases may arise:
One is like 1984, where the last few terms are positive, and can be reduced by 1, without increasing the number of terms; the other is like 1985, where the last few terms are negative, and subtracting 1 will result in $-2 \cdot 3^{i} \quad(i=0,1,2,3)$, and since
$$
-2 \cdot 3^{i}=-3^{i+1}+3^{i}
$$

replacing $-2 \cdot 3^{i}$ with $-3^{i+1}+3^{i}$, although it adds one term, since the expression for 1985 has only 7 terms, adding one term will not exceed 8 terms, and when $i+1=4$ or 7, it will exactly cancel out with the positive terms in front, thus keeping the number of terms unchanged.
Therefore, for $k \in[0,1985]$, all numbers can be represented as
$$
k=\sum_{i=1}^{8} a_{i} 3^{i-1}, \quad a_{i} \in\{-1,0,1\} \text { . }
$$

If the $a_{i}$ in the formula are replaced with their opposites, then for $k \in[-1985,0]$, the above formula can also be used.
Thus, the required $n_{i}=3^{i-1}, \quad i=1,2, \cdots, 8$.