Three, 17 football fans plan to go to Cao Guo to watch the World Cup football matches, and they have selected a total of 17 matches. The booking of tickets meets the following conditions:
(i) Each person can book at most one ticket per match;
(ii) The tickets booked by any two people have at most one match in common;
(iii) Only one person has booked 6 tickets.

How many tickets can these fans book at most? Explain your reasoning.
Solution: Draw a $17 \times 17$ grid, with 17 columns representing 17 matches and 17 rows representing 17 people. If the $i$-th person has booked a ticket for the $j$-th match, then the center of the cell at the intersection of the $i$-th row and the $j$-th column is marked with a red dot. Thus, the problem is transformed into finding the maximum number of red dots in the grid such that no four red dots form the vertices of a rectangle with sides parallel to the grid lines, given that one row has 6 red dots.

Assume the first row has red dots in the first 6 cells.

Divide the $17 \times 17$ grid into two parts: a $17 \times 6$ and a $17 \times 11$. In the first part, the first row has 6 red dots, so the other 16 rows can have at most 1 red dot each. Therefore, this part can have at most 22 red dots. In the second part, the first row has no red dots, so we need to determine the maximum number of red dots in a $16 \times 11$ grid.

Let $x_i$ be the number of red dots in the $i$-th row, and consider a pair of red dots in the same row as a "red dot pair." Thus, the $i$-th row generates $\mathrm{C}_{x_i}^{2}$ red dot pairs (where $\mathrm{C}_{1}^{2}=\mathrm{C}_{0}^{2}=0$). Since no four red dots can form a rectangle with sides parallel to the grid lines, we have:
$$
\mathrm{C}_{x_1}^{2} + \mathrm{C}_{x_2}^{2} + \cdots + \mathrm{C}_{x_{16}}^{2} \leq \mathrm{C}_{11}^{2} = 55.
$$

It is easy to see that the sum on the left is minimized when $x_1, x_2, \cdots, x_{16}$ are as evenly distributed as possible (differing by at most 1). Therefore, when $x_1, x_2, \cdots, x_{16}$ include two 4s and fourteen 3s, $\mathrm{C}_{x_1}^{2} + \mathrm{C}_{x_2}^{2} + \cdots + \mathrm{C}_{x_{16}}^{2} = 54$, and $x_1 + x_2 + \cdots + x_{16} = 50$. It is clear that if the grid has 51 red dots, it cannot meet the requirements. Thus, the $17 \times 17$ grid can have at most 72 red dots.

In the following grid, there are 71 red dots, with the first 6 cells in the first row being red dots, and no four red dots form a rectangle with sides parallel to the grid lines. This indicates that the maximum number of red dots is at least 71.

The key is whether 72 red dots can meet the requirements. Suppose there are 72 red dots that meet the requirements. Then, the first 6 columns have 22 red dots, and the last 11 columns have 50 red dots. From the inequality and subsequent derivation, the distribution of 50 red dots in the $16 \times 11$ grid can only be one of two scenarios:
(1) Two rows each have 4 red dots, and the other 14 rows each have 3 red dots;
(2) Three rows each have 4 red dots, one row has 2 red dots, and the other 12 rows each have 3 red dots.

First, consider (1). In the $17 \times 17$ grid, let the red dots in the first row be numbered 1, 2, 3, 4, 5, 6, and the red dots in the second row be numbered 1, 7, 8, 9, 10. The third row also has 5 red dots, and the last 14 rows each have 4 red dots.

Examine the distribution of red dots in columns 7, 8, 9, 10. In this case, each of the last 15 rows can have at most 1 red dot in these 4 columns. If a row has 1 red dot, then the number of red dots in the last 7 columns of that row is 2 or 3 (only one row has 3). Since each of these 4 columns can form 7 different "red dot pairs" with the last 7 columns, each column can have at most 3 red dots in the last 15 rows, so the $17 \times 4$ grid can have at most 16 red dots.

Removing the first 2 rows and the first 10 columns, at most 38 red dots are removed, leaving at least 34 red dots in the $15 \times 7$ grid. Since $34 = 3 \times 4 + 2 \times 11$, these red dots form at least
$$
3 \times 4 + 11 = 23
$$
different "red dot pairs," and $23 > 21 = \mathrm{C}_{7}^{2}$, which leads to a contradiction.

Next, consider (2). Let the red dots in the first row be numbered 1, 2, 3, 4, 5, 6, the red dots in the second row be numbered 1, 7, 8, 9, 10, the third and fourth rows each have 5 red dots, and the last row has 3 red dots, with the other 12 rows each having 4 red dots.

Again, examine the distribution of red dots in columns 7, 8, 9, 10. If there are at most 16 red dots, it leads to a contradiction as in (1). However, since the last row has only 3 red dots, one of which is in the first 6 cells. If none of the cells 7, 8, 9, 10 have a red dot, it is the same as the previous case; if one of the cells 7, 8, 9, 10 has a red dot, then the last 7 cells have only one red dot, leading to 17 red dots in the $17 \times 4$ grid. Removing the last row, the remaining $14 \times 7$ grid has 32 red dots. Since $32 = 3 \times 4 + 2 \times 10$, the number of "red dot pairs" is at least
$$
3 \times 4 + 10 = 22 > 21 = \mathrm{C}_{7}^{2}.
$$
This leads to a contradiction.

In conclusion, the maximum number of tickets that 17 people can book is 71.