Three. (50 points) The most recent mathematics competition consisted of 6 problems, with each correct answer scoring 7 points and each incorrect (or unanswered) question scoring 0 points. After the competition, a certain team scored a total of 161 points, and it was found during the score tallying that: any two participants in the team had at most two problems answered correctly in common, and no three participants answered the same two problems correctly. How many participants are there in the team at least?



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We find that by moving a point in the first column to another column in the same row, we can reduce the number of columns in Figure 6. For example, by making the move $(6,1) \rightarrow(6,2)$, we can simultaneously make the moves $(4,10) \rightarrow(6,3)$, $(3,9) \rightarrow(6,4)$, and $(5,9) \rightarrow(6,7)$, thus obtaining Figure 7 with 23 red points. Similarly, we can obtain Figure 8. This indicates that the minimum value of $n$ is no more than 7.

Three, suppose the team has $n$ players, denoted as $a_{1}$, $a_{2}$, ..., $a_{n}$, and the 6 problems are numbered as $1, 2, \cdots, 6$. We create a $6 \times n$ grid table, where if player $a_{i} (i=1,2, \cdots, n)$ answers the $j$-th problem $(j=1,2, \cdots, 6)$ correctly, we color the center of the small grid $(j, i)$ in the $j$-th row and $i$-th column red. Our problem is to find the minimum value of $n$ such that in a $6 \times n$ grid table, there are at least 23 red points without any "horizontal" 6 points.

If the first column has 6 red points, then the subsequent columns can have at most 2 red points each. Since $C_{6}^{2}=15>9$, we can take columns 2 to 10, where columns 2 to 9 each have 2 red points, and column 10 has 1 red point (as shown in Figure 6), which satisfies the problem's conditions. This indicates that the minimum value of $n$ is no more than 10.
Next, we prove that the minimum value of $n$ is greater than 6.
For a grid table with exactly 6 columns, by the pigeonhole principle, at least one column must have no fewer than 4 red points. Without loss of generality, assume the first column, and the first 4 rows of the first column have red points. If a column has 2 red points, we call it a "row pair" in that column. Thus, in the first 4 rows, each of the 5 columns other than the first column can have only one row pair. Therefore, the first 4 rows have a total of $C_{4}^{2}+5=11$ row pairs. Consider the last two rows: if the first column has additional red points, then the row with red points cannot have any other red points. If the first column has 2 additional red points, this can add 9 row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+9=20$; if the first column has 1 additional red point, assume the 5th row of the first column has a red point, then even if the 6th row has red points in all other columns except the first column, this can add $C_{4}^{1}+5 \times 2=14$ row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+14=25$; if the first column has no additional red points, then in the last two rows, there can be at most 2 row pairs, which occupy two columns, and in the remaining three columns, each column can have at most 1 red point, thus adding $2 \times 5+3 \times 2=16$ row pairs, making the total number of row pairs in the $6 \times 6$ grid table $11+16=27$. This indicates that 27 is the maximum possible total number of row pairs.

Let the number of red points in the $i$-th column be $x_{i} (i=1,2, \cdots, 6)$, and $\sum_{i=1}^{6} x_{i}=k$. Then the total number of row pairs is $\sum_{i=1}^{6} C_{x_{i}}^{2} \leqslant 27$, i.e.,
$$
\sum_{i=1}^{6} x_{i}^{2}-\sum_{i=1}^{6} x_{i} \leqslant 54.
$$

By the Cauchy-Schwarz inequality, we have
$$
\sum_{i=1}^{6} x_{i}^{2} \geqslant \frac{1}{6}\left(\sum_{i=1}^{6} x_{i}\right)^{2}=\frac{1}{6} k^{2}.
$$

Therefore, $\frac{k^{2}}{6} \leqslant k+54$.
Solving this, we get $3-3 \sqrt{37} \leqslant k \leqslant 3+3 \sqrt{37}$.
Since $k$ is a positive integer, $k \leqslant 21$. This indicates that the maximum number of red points in a $6 \times 6$ grid table is 21.

Furthermore, when $n \leqslant 5$, the total number of red points in the grid table is no more than $4 \times 5=20$. This indicates that the minimum value of $n$ is at least 7. In summary, the team must have at least 7 players.
(Liu Shixiong, Zhongshan Affiliated High School of South China Normal University, 528447)