18. (IMO-28 Preliminary Question) 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 $a_{1}, a_{2}, \cdots, a_{8}$, where each $a_{i}$ belongs to the set $\{-1,0,1\}$, such that $k=\sum_{i=1}^{8} a_{i} n_{i}$.
18. Consider the ternary representation of 1985: $1985=2 \cdot 3^{6}+2 \cdot 3^{5}+3^{3}+3^{2}+3+2$.

However, according to the problem, the coefficients in the representation cannot be 2, and since $2=3-1$, 1985 can be expressed as $1985=3^{7}-3^{5}+3^{4}-3^{3}-3^{2}-3-1$. Similarly, we have $1984=3^{7}-3^{5}+3^{3}+3^{2}+3+1$.
Starting from 1984 and subtracting 1 step by step, two scenarios may occur:
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 $-2 \cdot 3^{j} (i=0,1,2,3)$. 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, the expression for 1985 has only 7 terms, so adding one term will not exceed 8 terms. 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 $k \in [0,1985]$, each number can be represented as $k=\sum_{i=1}^{k} a_{i} 3^{i-1}, a_{i} \in \{-1,0,1\}$.
If the $a_{i}$ in the expression are replaced with their opposites, then for $k \in [-1985,0]$, the number can also be represented by the above formula.
Thus, the required $n_{i}=3^{i-1}, i=1,2, \cdots, 8$.