6. Solution: Let these n points be A1,A2,⋯,An. First, consider an example:
If n is odd, connect the segments A1A2,A1A3,⋯,A1An;A2A3,A4A5,⋯,An−1An.
If n is even, connect the segments A1A2,A1A3,⋯,A1An;A2A3,A4A5,⋯,An−2An−1,A2An.
Clearly, the segments connected by the above method satisfy the conditions, so the minimum value of m is less than or equal to [23n−2]. (Note: [23n−2] is the number of segments connected in the above method, where [x] denotes the greatest integer not exceeding x.)
Next, we prove that these n points require at least [23n−2] segments to meet the requirements of the problem.
In fact, if each of A1,A2,⋯,An is connected to at least 3 segments, then m⩾23n>[23n−2];
If one of them (let's say A1) is connected to no more than 2 segments, there are two cases:
(1) A1 is connected to only 1 segment, say A1A2, then there is no point that is connected to both A1 and A2, which is a contradiction. Similarly, if A1 is not connected to any segment, it also leads to a contradiction.
(2) A1 is connected to exactly 2 segments, A1A2 and A1A3. In this case, the segment A2A3 must be connected (because for the two points A1 and A2, only A3 can be connected to both A1 and A2).
Considering the points A1 and Ai(i⩾4), Ai must be connected to either A2 or A3. Also, from (1), Ai must be connected to at least 2 segments, so the number of segments connected from A4,A5,⋯,An (a total of n−3 points) is at least n−3+[2n−2]. Therefore, the total number of segments connected from these n points is at least 3+(n−3)+[2n−2]=[23n−2].
Thus, the minimum value of m is [23n−2].