Solution: First, prove that $|B| \geq 17$. Consider that when all elements in $A$ are changed to their opposites, the set $B$ remains unchanged, so we can assume without loss of generality that the number of positive numbers in $A$ is no less than the number of negative numbers. We will discuss the following cases:
Case one: $A$ contains no negative numbers.
Let $a_{1} < a_{2} < \cdots < a_{10}$, and $b_{1} < b_{2} < \cdots < b_{6} < 0 < c_{1} < c_{2} < \cdots < c_{l}$. Then
$$
a_{1} a_{2} c_{1} b_{2} > \cdots > c_{1} b_{k} > c_{2} b_{k} > \cdots > c_{l} b_{k},
$$

These are $k+l-1=10$ elements in $B$, and they are non-positive; also,
$$
b_{2} b_{3} < b_{2} b_{4} < b_{2} b_{5} < b_{2} b_{6} < b_{3} b_{6} < b_{4} b_{6} < b_{5} b_{6},
$$

These are 7 elements in $B$, and they are positive. Therefore, $|B| \geq 10+7=17$.
Thus, $|B| \geq 17$. $\qquad$
On the other hand, let $A=\left\{0, \pm 1, \pm 2, \pm 2^{2}, \pm 2^{3}, \pm 2^{4}\right\}$, then
$$
B=\left\{0, -1, \pm 2, \pm 2^{2}, \pm 2^{3}, \ldots, \pm 2^{6}, \pm 2^{7}, -2^{8}\right\}
$$

is a 17-element set.
In conclusion, the minimum number of elements in $B$ is 17.
50 points
Four, (This question is worth 50 points) Let $A$ be any set of 11 real numbers. Let the set $B=\{uv \mid u, v \in A, u \neq v\}$. Find the minimum number of elements in $B$.

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