Problem 1761

[Solution] (1) Let the mass numbers of these $k$ weights be $a_{1}, a_{2}, \cdots, a_{k}$, and $1 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{k}, a_{i} \in \mathbb{Z}, 1 \leqslant i \leqslant k$. Since weights can be placed on both sides of the balance, the mass that can be measured is
$\sum_{i=1}^{k} x_{i} a_{i}, x_{i} \in\{-1,0,1\}$.
If using these $k$ weights can measure the mass of items that are $1,2, \cdots, n$, then the above representation includes $1,2, \cdots, n$. By symmetry, it is easy to see that it also includes $-1,-2, \cdots,-n$. Therefore,
$$\left\{\sum_{i=1}^{k} x_{i} a_{i} \mid x_{i} \in\{-1,0,1\}\right\} \supseteq\{0, \pm 1, \pm 2, \cdots, \pm n\}$$

Since the number of elements in the set $\left\{\sum_{i=1}^{k} x_{i} a_{i} \mid x_{i} \in\{-1,0,1\}\right\}$ does not exceed $3^{k}$, and the set $\{0, \pm 1, \pm 2, \cdots, \pm n\}$ contains $2 n+1$ elements, therefore
$$\begin{array}{l} 3^{k} \geqslant 2 n+1 \\ n \leqslant \frac{3^{k}-1}{2} \end{array}$$

If $\frac{3^{m-1}-1}{2}<n \leqslant \frac{3^{m}-1}{2},(m \geqslant 1, m \in \mathbb{Z})$, then $k \geqslant m$.
$$\text { On the other hand, if } \frac{3^{m-1}-1}{2}<n \leqslant \frac{3^{m}-1}{2},(m \geqslant 1, m \in \mathbb{Z}) \text {, then we can }$$

take $m$ weights: $a_{1}=1, a_{2}=3, \cdots, a_{m}=3^{m-1}$ to measure all items with mass $1,2, \cdots, n$ grams.

In fact, by the ternary representation of numbers, for any $0 \leqslant p \leqslant 3^{m}-1$, there exist $y_{i} \in\{0,1,2\}, 1 \leqslant i \leqslant m$, such that
$$p=\sum_{i=1}^{\prime \prime} y_{i} 3^{i-1}$$

Thus, $p-\frac{3^{m}-1}{2}=\sum_{i=1}^{m} y_{i} \cdot 3^{i-1}-\sum_{i=1}^{m} 3^{i-1}$
$$=\sum_{i=1}^{m}\left(y_{i}-1\right) 3^{i-1}$$

Let $l=p-\frac{3^{m}-1}{2}$, then $-\frac{3^{m}-1}{2} \leqslant l \leqslant \frac{3^{m}-1}{2}$;
Let $x_{i}=y_{i}-1$, then $x_{i} \in\{-1,0,1\}$, thus, we have
$$l=\sum_{i=1}^{m} x_{i} \cdot 3^{i-1}$$

Since $n \leqslant \frac{3^{m}-1}{2}$, for any $l \in\{1,2, \cdots, n\}$, there exist $x_{i}, 1 \leqslant i \leqslant m$, such that
$$l=\sum_{i=1}^{m} x_{i} \cdot 3^{i-1}=\sum_{i=1}^{m} x_{i} a_{i}$$

This means that using weights with mass $1,3, \cdots, 3^{m-1}$, we can measure all items with mass $1,2, \cdots, n$ grams, where $n \leqslant \frac{3^{m}-1}{2}$.

In summary, the minimum value of $k$ is $f(n)=m$, where $m$ satisfies the inequality $\frac{3^{m-1}-1}{2}<n \leqslant \frac{3^{m}-1}{2}$.
(2) First, prove that when $\frac{3^{m-1}-1}{2}<n<\frac{3^{m}-1}{2}$, the composition of $f(n)$ weights, besides the one already mentioned in (1):
$$\begin{array}{l} a_{1}=1, a_{2}=3, \cdots, a_{m}=3^{m-1} \text {, there is at least one more way: } \\ a_{1}=1, a_{2}=3, \cdots, a_{m-1}=3^{m-2}, a_{m}=3^{m-1}-1 . \end{array}$$

In fact, if $1 \leqslant l \leqslant \frac{3^{m-1}-1}{2}$, then by (1), there exist $x_{i} \in\{-1,0,1\}$, such that
$$\begin{array}{l} l=\sum_{i=1}^{m-1} x_{i} \cdot 3^{i-1} \\ \quad=\sum_{i=1}^{m-1} x_{i} \cdot 3^{i-1}+0 \cdot\left(3^{m-1}-1\right) . \\ \text { If } \frac{3^{m-1}-1}{2}<l \leqslant n<\frac{3^{m}-1}{2} \text {, then } l+1 \leqslant \frac{3^{m}-1}{2}, \end{array}$$

Thus, by (1), there exist $x_{i} \in\{-1,0,1\}$, such that
$$l+1=\sum_{i=1}^{m} x_{i} \cdot 3^{i-1}$$

And it must be that $x_{m}=1$. Thus,
$$l=\sum_{i=1}^{m-1} x_{i} \cdot 3^{i-1}+1 \cdot\left(3^{m-1}-1\right)$$

Therefore, using weights $a_{1}=1, a_{2}=3, \cdots, a_{m-1}=3^{m-2}, a_{m}=3^{m-1}-1$ can measure all items with mass $1,2, \cdots, n$ grams. Thus, in this case, the composition of $f(n)$ weights is not unique.

Next, prove that when $n=\frac{3^{m}-1}{2}$, the composition of $f(n)$ weights is unique, i.e., $a_{i}=3^{i-1}(1 \leqslant i \leqslant m)$.

In fact, if $m$ weights $a_{1}, a_{2}, \cdots, a_{m}$ can measure all items with mass $1,2, \cdots, n=\frac{3^{m}-1}{2}$ grams, then for each $-\frac{3^{m}-1}{2} \leqslant l \leqslant \frac{3^{m}-1}{2}$, there exist
$$l=\sum_{i=1}^{m} x_{i} a_{i}, x_{i} \in\{-1,0,1\}$$

Therefore,
$$\left\{\sum_{i=1}^{m} x_{i} a_{i} \mid x_{i} \in\{-1,0,1\}\right\} \supseteq\left\{0, \pm 1, \cdots, \pm \frac{3^{m}-1}{2}\right\}$$

Notice that the left-hand side set contains at most $3^{m}$ elements, and the right-hand side set contains exactly $3^{m}$ elements, thus, we have
$$\left\{\sum_{i=1}^{m} x_{i} a_{i} \mid x_{i} \in\{-1,0,1\}\right\}=\left\{0, \pm 1, \cdots, \pm \frac{3^{m}-1}{2}\right\}$$

And $\sum_{i=1}^{m} a_{i}=\frac{3^{m}-1}{2}$
Adding $\frac{3^{m}-1}{2}$ to each element of the set, we get
$$\left\{\sum_{i=1}^{m} x_{i} a_{i}+\sum_{i=1}^{m} a_{i} \mid x \in\{-1,0,1\}\right\}=\left\{0,1,2, \cdots, 3^{m}-1\right\}$$

That is, $\left\{\sum_{i=1}^{m} y_{i} a_{i} \mid y_{i} \in\{0,1,2\}\right\}=\left\{0,1,2, \cdots, 3^{m}-1\right\}$.
And for each $l\left(0 \leqslant l \leqslant 3^{m}-1\right)$, it can be uniquely represented as $l=\sum_{i=1}^{m} y_{i} a_{i}$. Assume $a_{1}<a_{2}<\cdots<a_{m}$. Therefore, $a_{1}$ is the smallest positive integer in the set $\left\{0,1,2, \cdots, 3^{m}-1\right\}$, i.e., $a_{1}=1$. Assume $a_{1}=1, a_{2}=3, \cdots, a_{s}=3^{s-1}$. Then