10. 12 people gather for dinner at the same restaurant every week, each time divided into three tables, with four people at each table. It is required that any two people sit at the same table at least once. Question: What is the minimum number of weeks such dinners need to be held?
10. Since the number of unique "2-person pairs" is $C_{12}^{2}=66$, and each dinner only produces $3 C_{4}^{2}=18$ "2-person pairs", it is necessary to have $\left[\frac{66}{18}\right]+1=4$ weeks. If it is exactly 4 weeks, since each table has 4 people in the first week, in the following weeks, there must be two people sitting at the same table (because there are only 3 tables each week, and 4 people can form $C_{4}^{2}=6$ "2-person pairs"), so the number of unique "2-person pairs" in 4 weeks is at most $4 \times 18-3 \times 3=63$, which contradicts the requirement of generating 66 unique "2-person pairs". Therefore, it must take at least 5 weeks. Below is an example of a feasible 5-week schedule (using 1, $2, \cdots, 12$ to represent people): 
- Week 1: $(1,2,3,4),(5,6,7,8),(9,10,11,12)$; 
- Week 2: $(1,2,5,6),(3,4,9,10),(7,8,11,12)$; 
- Week 3: $(1,2,7,8),(3,4,11,12),(5,6,9,10)$; 
- Week 4: $(1,2,9,10),(3,4,7,8),(5,6,11,12)$; 
- Week 5: $(1,2,11,12),(3,4,5,6),(7,8,9,10)$.