14. (15 points) Math Competition, 8 fill-in-the-blank questions, 4 points for each correct answer, 0 points for each incorrect answer; 6 short-answer questions, 7 points for each correct answer, 0 points for each incorrect answer, 400 participants, what is the minimum number of people who have the same total score?
【Analysis】First, find out how many different outcomes there are, then divide 400 by the number of groups to see if there is any remainder, which is the maximum value of the average distribution.

【Solution】Solution: Method one: Let the number of 4-point questions answered correctly be $a$, and the number of 7-point questions answered correctly be $b$, then $a$ can take 9 values from 0 to 8, and $b$ can take 7 values from 0 to 6,

The number of scoring situations is $9 \times 7=63$, then consider the repeated scoring situations, when $a^{\prime}=a+7, b^{\prime}=b-4$, the scores are the same, i.e., $(a, b)=(0,6)$ and $(7,2),(0,5)$ and $(7,1),(0,4)$ and $(7,0),(1,6)$ and $(8,2),(1,5)$ and $(8,1),(1,4)$ and $(8,0)$; in 6 cases, the scores will be the same.

Therefore, there are $63-6=57$ (types) of different scores, $400 \div 57=7 \cdots 1.7+1=8$, at least 8 people have the same score, so the answer is: 8

Method two: According to the problem:
8 fill-in-the-blank questions and 6 answer questions total $8 \times 4+6 \times 7=74$ (points)
When no answer questions are answered correctly: There are 9 situations: $0,4,8,12,16,20,24,28,32$.
When 1 answer question is answered correctly: There are 9 situations: $7,11,15,19,23,27,31,35,39$.
When 2 answer questions are answered correctly: There are 9 situations: $14,18,22,26,30,34,38,42,46$.
When 3 answer questions are answered correctly: There are 9 situations: $21,25,29,33,37,41,45,49,53$.
When 4 answer questions are answered correctly: There are 9 situations: $28,32,36,40,44,48,52,56,60$. 2 repeats, total 7.
When 5 answer questions are answered correctly: There are 9 situations: $35,39,43,47,51,55,59,63,67$. 2 repeats, total 7.
When 6 answer questions are answered correctly: There are 9 situations: $42,46,50,54,58,62,66,70,74$. 2 repeats, total 7. There are $4 \times 9+7 \times 3=57$.
$$
400 \div 57=7 \cdots 1.7+1=8 \text {. }
$$

So the answer is: 8.