Three. (25 points) There are 20 weights, all of which are integers, such that any integer weight $m(1 \leqslant m \leqslant 2002)$ can be balanced by placing it on one pan of a scale and some of the weights on the other pan. What is the smallest possible value of the heaviest weight among these 20 weights?
Let's assume the weights of these 20 weights are \(a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{20}\), where \(a_{i} (1 \leqslant i \leqslant 20)\) are positive integers. It is easy to see that
\[
a_{1}=1, a_{k+1} \leqslant a_{1}+a_{2}+\cdots+a_{k}+1 (1 \leqslant k \leqslant 19).
\]
\[
\text{Then } a_{2} \leqslant 2, a_{3} \leqslant 4, a_{4} \leqslant 8, a_{5} \leqslant 16, a_{6} \leqslant 32,
\]
\[
\begin{array}{l}
a_{7} \leqslant 64, a_{8} \leqslant 128. \\
\text{Also, since } a_{1}+a_{2}+\cdots+a_{20} \geqslant 2002, \\
\therefore 12 a_{20} \geqslant a_{9}+a_{10}+\cdots+a_{20},
\end{array}
\]
or \(12 a_{20} \geqslant (a_{1}+a_{2}+\cdots+a_{20}) - (a_{1}+a_{2}+\cdots+a_{8})\)
\[
\geqslant 2002 - (1+2+4+\cdots+128) \geqslant 1747.
\]

Thus, \(a_{20} \geqslant 145 \frac{7}{12}\).
Since \(a_{20}\) is an integer, \(\therefore a_{20} \geqslant 146\).
Let \(a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, a_{6}=32\). \(a_{7}=64, a_{8}=128, a_{9}=a_{10}=a_{11}=a_{12}=a_{13}=145, a_{14}\)
\[
=a_{15}=\cdots=a_{20}=146.
\]
Using \(a_{1}, a_{2}, \cdots, a_{8}\) these 8 weights, we can measure integer weights from 1 to 255. Adding \(a_{9}\), we can measure integer weights from 1 to 400. Adding \(a_{10}\), we can measure integer weights from 1 to 545. And so on, 20 weights can measure integer weights from 1 to 2002.
Therefore, the minimum value of the heaviest weight is 146.
(Yang Jin, No. 13 Middle School, Wuhu City, Anhui Province, 241002)