Three, (50 points) Try to find the smallest possible value of the positive integer $k$, such that the following proposition holds: For any $k$ integers $a_{1}, a_{2}, \cdots, a_{k}$ (equality is allowed), there must exist corresponding $k$ integers $x_{1}, x_{2}, \cdots, x_{k}$ (equality is also allowed), and $\left|x_{i}\right| \leqslant 2(i=1,2, \cdots, k),\left|x_{1}\right|+\left|x_{2}\right|+\cdots+$ $\left|x_{k}\right| \neq 0$, such that 2003 divides $x_{1} a_{1}+x_{2} a_{2}+\cdots$ $+x_{k} a_{k}$.
Three, first prove that the proposition holds when $k=7$. For this, consider the sum
$$
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)=y_{1} a_{1}+y_{2} a_{2}+\cdots+y_{7} a_{7} \text {, }
$$

where $y_{i} \in\{-1,0,1\}$.
There are $3^{7}=2187$ such sums, and since $2187>2003$, by the pigeonhole principle, there must be two different sums $S\left(y_{1}, y_{2}, \cdots, y_{7}\right)$ and $S\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots, y_{7}^{\prime}\right)$ that have the same remainder when divided by 2003.
Therefore, 2003 divides
$$
\begin{array}{l}
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)-S\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots, y_{7}^{\prime}\right) \\
=\left(y_{1}-y_{1}^{\prime}\right) a_{1}+\left(y_{2}-y_{2}^{\prime}\right) a_{2}+\cdots+\left(y_{7}-y_{7}^{\prime}\right) a_{7} .
\end{array}
$$

where $y_{i}-y_{i}^{\prime} \in\{-2,-1,0,1,2\}, i=1,2, \cdots, 7$ (since $\left.y_{i} 、 y_{i} \in\{-1,0,1\}, i=1,2, \cdots, 7\right)$, and at least one $y_{i}-y_{i}^{\prime} \neq 0$ (because the arrays $\left(y_{1}, y_{2}, \cdots, y_{7}\right) \neq\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots\right.$, $\left.y_{7}^{\prime}\right)$ ). At this point, taking $x_{i}=y_{i}-y_{i}^{\prime}, i=1,2, \cdots, 7$, satisfies the requirement.
Thus, the proposition holds when $k=7$.
Next, prove that the proposition does not hold when $k=6$. For this, we provide a counterexample.

Take $a_{1}=3^{1}, a_{2}=3^{2}, a_{3}=3^{3}, a_{4}=3^{4}, a_{5}=3^{5}, a_{6}=$ $3^{6}$, then for any 6 integers $x_{1}, x_{2}, \cdots, x_{6},\left|x_{i}\right| \leqslant$ $2,\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{6}\right| \neq 0$, the sum
$$
S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{6} x_{6}
$$

is always a multiple of 3.
$$
\begin{array}{l}
\text { Then } S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \text { | } \\
\leqslant 2\left(3^{1}+3^{2}+\cdots+3^{6}\right) \\
=3\left(3^{6}-1\right)=3\left(3^{3}-1\right)\left(3^{3}+1\right) \\
=3 \times 26 \times 28=3 \times 728 .
\end{array}
$$

Assume without loss of generality that the largest index $k_{1}$ of the non-zero elements $x_{1}, x_{2}, \cdots, x_{6}$ is $x_{k_{1}}$, i.e., $x_{k_{1}} \neq 0$, and $x_{k_{1}+1}=x_{k_{1}+2}=\cdots=x_{6}=$ 0, then
$$
\begin{array}{l}
S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{k_{1}} x_{k_{1}} \\
=3^{1} x_{1}+3^{2} x_{2}+\cdots+3^{k_{1}} x_{k_{1}} .
\end{array}
$$

Additionally, assume without loss of generality that $x_{k_{1}}>0$ (if $x_{k_{1}}<0$, the argument is similar).
If $k_{1} \geqslant 2$, then

If $k_{1}=1$, then
$$
x_{1} \neq 0, S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}>0 \text {. }
$$

In summary,
$$
\begin{array}{l}
3 \mid S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \text {, and } S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \neq 0 \text {, } \\
\left|S\left(x_{1}, x_{2}, \cdots, x_{6}\right)\right| \leqslant 3 \times 728 .
\end{array}
$$

Clearly, 2003 and 3 are coprime.
If 2003 divides $S\left(x_{1}, x_{2}, \cdots, x_{6}\right)$, then $S\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{6}\right)=3 t$, where $t$ is an integer, and $1 \leqslant|t| \leqslant 728$. Thus, $2003 \mid t$, which contradicts $1 \leqslant|t| \leqslant 728$. Therefore, when taking $a_{i}=3^{i}$, $i=1,2, \cdots, 6$, it is impossible to have $x_{1}, x_{2}, \cdots, x_{6} \in \mathbf{Z},\left|x_{i}\right|$ $\leqslant 2, i=1,2, \cdots, 6,\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{6}\right| \neq 0$, such that
$2003$ divides $S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{6} x_{6}$. This counterexample shows that the proposition does not hold when $k=6$.
From the above two steps, the smallest positive integer $k$ is 7.
(Wu Weizhao, Department of Mathematics, College of Science, Guangzhou University, 510405)