6.12 friends have a weekly dinner together, each week they are divided into three groups, each group 4 people, and different groups sit at different tables. If it is required that any two of these friends sit at the same table at least once, then at least how many weeks are needed.
6.5.

First, for any individual, sitting with 3 different people each week, it would take at least 4 weeks.

Second, there are $C_{12}^{2}=66$ pairs among 12 people. Each table has $\mathrm{C}_{4}^{2}=6$ pairs, so in the first week, $3 \times 6=18$ pairs get to know each other.

Since 4 people sit at 3 tables, after the first week, at least two people at each table must have sat together in the first week. Therefore, the maximum number of new pairs that can form each week is $6-1=5$ pairs per table, totaling 15 pairs.
Since $18+15+15+15=63$, it is impossible to have all 66 pairs sit together in 4 weeks, so it must take 5 weeks.

Using 5 weeks is feasible. For example, 18 pairs in the first week, and 12 pairs each in the remaining 4 weeks, totaling $18+12 \times 4=66$ pairs. Below is a specific seating arrangement:
\begin{tabular}{crrrrrrrrrrrrrr} 
Week & & \multicolumn{5}{c}{ Table 1} & & & & Table 2 & & & \multicolumn{5}{c}{ Table 3} & \\
1 & 1 & 2 & 3 & 4 & 5 & 6 & 9 & 10 & 7 & 8 & 11 & 12 \\
2 & 1 & 2 & 5 & 6 & 3 & 4 & 7 & 8 & 9 & 10 & 11 & 12 \\
3 & 1 & 2 & 7 & 8 & 3 & 4 & 9 & 10 & 5 & 6 & 11 & 12 \\
4 & 1 & 2 & 9 & 10 & 3 & 4 & 11 & 12 & 5 & 6 & 7 & 8 \\
5 & 1 & 2 & 11 & 12 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10
\end{tabular}