Three. (20 points) A certain mathematics competition had a total of 15 questions. The table below shows the statistics for the number of people who got $n (n=0,1,2, \cdots, 15)$ questions correct.
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline$n$ & 0 & 1 & 2 & 3 & $\cdots$ & 12 & 13 & 14 & 15 \\
\hline Number of people who got $n$ questions correct & 7 & 8 & 10 & 21 & $\cdots$ & 15 & 6 & 3 & 1 \\
\hline
\end{tabular}

If it is also known that the students who got 4 questions and more than 4 questions correct each averaged 6 questions correct, and the students who got 10 questions and 10 questions or fewer correct each averaged 4 questions correct. How many people does this table at least include?
Three, Solution 1 From the statistical table, we know: The total number of people who got $0-3$ questions correct is $7+8+10+21=40$ people, and the total number of questions they got correct is $7 \times 0+8 \times 1+10 \times 2+21 \times 3=91$ (questions); The total number of people who got $12-15$ questions correct is $15+6+3+1=25$ (people), and the total number of questions they got correct is $15 \times 12+6 \times 13+3 \times 14+1$ $\times 15=315$ (questions).

Let $x_{0}, x_{1}, \cdots, x_{15}$ represent the number of people who got 0, 1, $\cdots, 15$ questions correct, respectively. From the problem, we have
$$
\frac{4 x_{4}+5 x_{5}+\cdots+15 x_{15}}{x_{4}+x_{5}+\cdots+x_{15}}=6,
$$

and $\frac{0 x_{0}+x_{1}+2 x_{2}+\cdots+10 x_{10}}{x_{0}+x_{1}+\cdots+x_{10}}=4$,
which means $4 x_{4}+5 x_{5}+\cdots+15 x_{15}=6\left(x_{4}+x_{5}+\cdots+x_{15}\right)$,
$$
0 x_{0}+x_{1}+2 x_{2}+\cdots+10 x_{10}=4\left(x_{0}+x_{2}+\cdots+x_{10}\right) \text {. }
$$

Subtracting the two equations, we get
$$
\begin{array}{l} 
11 x_{11}+12 x_{12}+\cdots+15 x_{15}-\left(x_{1}+2 x_{2}+3 x_{3}\right) \\
= 6\left(x_{4}+x_{5}+\cdots+x_{15}\right)-4\left(x_{0}+x_{1}+\cdots+x_{10}\right) \\
= 6\left(x_{11}+x_{12}+\cdots+x_{15}\right)-4\left(x_{0}+x_{1}+x_{2}+x_{3}\right) \\
+2\left(x_{4}+x_{5}+\cdots+x_{10}\right) \\
= 4\left(x_{11}+x_{12}+\cdots+x_{15}\right)-6\left(x_{0}+x_{1}+x_{2}+x_{3}\right) \\
+2\left(x_{0}+x_{1}+\cdots+x_{15}\right) \\
= 4 x_{11}+4\left(x_{12}+x_{13}+x_{14}+x_{15}\right)-6\left(x_{0}+x_{1}+x_{2}\right. \\
\left.\quad+x_{3}\right)+2\left(x_{0}+x_{1}+\cdots+x_{15}\right) .
\end{array}
$$

And we have already found that $x_{0}+x_{1}+x_{2}+x_{3}=46$,
$$
\begin{array}{l}
x_{12}+x_{13}+x_{14}+x_{15}=25, \\
0 x_{0}+x_{1}+2 x_{2}+3 x_{3}=91, \\
12 x_{12}+13 x_{13}+14 x_{14}+15 x_{15}=315,
\end{array}
$$

Substituting these into the above equation, we get
$$
\begin{array}{l}
11 x_{11}+315-91 \\
=4 x_{11}+4 \times 25-6 \times 46+2\left(x_{0}+x_{1}+\cdots+x_{15}\right) .
\end{array}
$$
$$
\text { Therefore, } x_{0}+x_{1}+\cdots+x_{15}=200+3.5 x_{11},\left(x_{11} \geqslant 0\right)
$$

Thus, when $x_{11}=0$, the total number of people counted in the statistics $x_{0}+x_{1}+\cdots+$ $x_{15}$ is at least 200 people.
Solution 2 (The four sums are given the same as in Solution 1)
Let this table have counted a total of $y$ people, and the number of people who got 11 questions correct is $x$ people. From the problem, we can derive the functional relationship between $y$ and $x$
$$
6(y-46)-315-4(y-x-25)+11 x-91 \text {, }
$$

Simplifying, we get $y=200+3.5 x(x \geqslant 0)$,
Therefore, when $x=0$, $y=200$ is the minimum.
Thus, this table has counted at least 200 people.