Five. (Full marks 30 points) There are several weights of 9 grams and 13 grams each. To weigh an object of 3 grams on a balance, what is the minimum number of such weights needed? Prove your conclusion.

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On the three pans of a balance, assuming that when the balance is level, 9 grams are used, and the number is an integer. Similarly, assuming that 13 grams of weights are used $|y|$ times. Therefore, when the balance is level and measures a 3-gram object, there should be
$$
9 x+13 y=3 \text {. }
$$

The problem becomes finding the minimum value of $|x|+|y|$.
First, we can take any integer solution of the equation, for example, because $y=$ $\frac{3-9 x}{13}$, when $x=9$, we get $y=-6$. Using this solution, we have
$$
\left\{\begin{array}{l}
9 \times 9-13 \times 6=3, \\
9 x+13 y=3 .
\end{array}\right.
$$

Subtracting the two equations, we get $9(x-9)+13(y+6)=0$,
$$
9(x-9)=-13(y+6) \text {. }
$$

Since 9 and 13 are coprime, $x-9$ must be divisible by 13. Therefore, let $x-9=13 k$, where $k$ is an integer. At this time, we have
$9 \times 13 k=-13(y+6),-(y+6)=9 k$.
In summary, we have $\left\{\begin{array}{l}x=9+13 k, \\ y=-6-9 k .\end{array} \quad k=0, \pm 1, \pm 2, \cdots\right.$
(i) When $k=0$, $x=9, y=-6$.
$$
|x|+|y|=15 \text {; }
$$
(ii) When $k \geqslant 1$, $|x| \geqslant 22,|y|>0$, thus
$$
|x|+|y|>22 \text {; }
$$
(iii) When $k \leqslant-1$, if $k=-1$, then $x=-4, y=3$, $|x|+|y|=7$; if $k12,|x|>0$, thus $|x|+|y|>12$.

From the above, we know that at least 7 such weights are needed, with 4 of them being 9 grams and 3 being 13 grams.
(Provided by Li Shuwen, No. 40 Middle School of Benzhuang City, Shandong Province)