Three, (50 points) Let $A=\{i \in \mathbf{N} \mid 1 \leqslant i \leqslant$ $2880\}, B \subseteq A,|B| \geqslant 9$. If all elements in set $A$ can be represented by the sum of no more than 9 different elements from $B$, find $\min |B|$, and construct a set corresponding to the minimum $|B|$.


Let $|B|=k \geqslant 9$.
According to the problem, we should have $S_{k}=\sum_{i=1}^{9} \mathrm{C}_{k}^{i} \geqslant 2880$.
Notice that $S_{9}=2^{9}-1, S_{10}=2^{10}-2$,
$$
\begin{array}{l}
S_{11}=2^{11}-2-\mathrm{C}_{11}^{10}=2035, \\
S_{12}=2^{12}-2-\mathrm{C}_{12}^{10}-\mathrm{C}_{12}^{11}=4016>2880 .
\end{array}
$$

Thus, $k \geqslant 12$.
Next, we prove:
$$
\begin{aligned}
B= & \left\{1,2,2^{2}, 2^{3}, 2^{4}, 2^{5}-1,2^{6}-2,2^{7}-4,\right. \\
& \left.2^{8}-8,2^{9}-16,2^{10}-32,2^{11}-80\right\} \\
= & \left\{b_{i} \mid i=0,1, \cdots, 11\right\}
\end{aligned}
$$

satisfies the conditions.
(1) First, we prove by mathematical induction: For any $n \in\left[1, b_{i}-1\right](i=5,6, \cdots, 10)$, $n$ can be expressed as the sum of at most $i-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, i-1\right\}$.

When $i=5$, for any $n \in\left[1, b_{5}-1\right]$, by binary knowledge,
$$
n=\sum_{j=0}^{4} \varepsilon_{j} j^{j}=\sum_{j=0}^{4} \varepsilon_{j} b_{j},
$$

where $\varepsilon_{j}=0$ or $1$, $\varepsilon_{j}$ are not all $1$, $j=0,1, \cdots, 4$.
That is, $n$ can be expressed as the sum of at most 4 different elements from $\left\{b_{j} \mid j=0,1, \cdots, 4\right\}$.
Assume the proposition holds for $i=k \geqslant 5$.
When $i=k+1$, by the induction hypothesis, it is easy to see that the proposition holds for $n \in \left[1, b_{k}\right]$; for $n \in\left[b_{k}+1, b_{k+1}-1\right]$, $n-b_{k} \in\left[1, b_{k}-1\right]$.

By the induction hypothesis, $n-b_{k}$ can be expressed as the sum of at most $k-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}$, hence $n$ can be expressed as the sum of at most $k$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k\right\}$.
(2) For $n \in\left[b_{10}+1, b_{11}-1\right]$, take $k \in \mathbf{N}$ such that
$$
\sum_{j=k}^{10} b_{j}>n \geqslant \sum_{j=k+1}^{10} b_{j} .
$$

If $k \leqslant 3$, then $n \geqslant \sum_{j=4}^{10} b_{j}=b_{11}+1$, which is a contradiction.
If $k=4$, then $n-\sum_{j=5}^{10} b_{j} \in\left[0, b_{4}-2\right]$, by (1) we know $n-\sum_{j=5}^{10} b_{j}$ can be expressed as the sum of at most 3 different elements from $\left\{b_{j} \mid j=0,1,2,3\right\}$. Hence $n$ can be expressed as the sum of at most 9 different elements from $\left\{b_{j} \mid j=0,1, \cdots, 10\}$.

If $k \geqslant 4$, then $n-\sum_{j=k+1}^{10} b_{j} \in\left[0, b_{k}-1\right]$, by (1) we know $n-\sum_{j=k+1}^{10} b_{j}$ can be expressed as the sum of at most $k-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}$. Hence $n$ can be expressed as the sum of at most
$$
(10-k)+(k-1)=9
$$

different elements from $\left\{b_{j} \mid j=0,1, \cdots, 10\}$.
(3) For $n \in\left[b_{11}+1,2880\right]$, then
$n-b_{11} \in\left[1, b_{10}-1\right]$.
Take $k \in \mathbf{N}$ such that $\sum_{j=k}^{9} b_{j}>n-b_{11} \geqslant \sum_{j=k+1}^{9} b_{j}$.
Thus, $n-b_{11}-\sum_{j=k+1}^{9} b_{j} \in\left[0, b_{k}-1\right]$.
By (1) we know $n-b_{11}-\sum_{j=k+1}^{9} b_{j}$ can be expressed as the sum of
$$
\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}
$$

of at most $k-1$ different elements.
Hence $n$ can be expressed as the sum of at most
$$
1+(9-k)+(k-1)=9
$$

different elements from $\left\{b_{j} \mid j=0,1, \cdots, 11\}$.
In summary, $\min |B|=12, B=\{1,2,4,8,16, 31,62,124,248,496,992,1968\}$ is a set that satisfies the conditions.