Let $ F_{1},...,F_{17}$ be fans. Each of them visit some match $ M_{1},...,M_{17}$ and say $ F_{1}$ visit 6 match's. Since every two visit at most 1 match in comon, we have \[ \sum_{i=1}^{17}{f_{i}\choose 2}\leq{17\choose 2}\] where $ f_{i}$ means number of games $ F_{i}$ visit. We are interested in maximum value of $ n =\sum_{i = 1}^{17}f_{i}$. By Jensen inequality we have: \[ {{1\over 16}(n-6)^{2}-(n-6)\over 2}\leq{17\choose 2}-{6\choose 2}\] So $ n\leq 76$. Now I can't find nor configuration for $ n = 76$ nor better uper bound.

I have a little improvement. Let $ F_{1}$ visit $ M_{1},...,M_{6}$. Then for all $ F_{i}' = F_{i}\setminus F_{1}$, $ i > 1$ condition: At most one match was same in the tickets booked by every two persons; is still valid. So we have now: \[ \sum_{i=2}^{17}{f_{i}'\choose 2}\leq{11\choose 2}\] beacuse all other games $ F_{i}'$ visit are at most in a set $ \{M_{7},...M_{17}\}$. Let $ n' =\sum_{i = 2}^{17}f_{i}'$. Then like before, applying Jensen, we get $ n'\leq 50$. This means: $ n\leq n'+16+6\leq 72$. Now i have configuration for $ n = 69$: $ F_{1}: 1,2,3,4,5,6$ $ F_{2}: 1,7,8,9,10$ $ F_{3}: 1,11,12,13,14$ $ F_{4}: 1,15,16,17$ $ F_{5}: 2,7,11,15$ $ F_{6}: 2,8,12,16$ $ F_{7}: 2,9,13,17$ $ F_{8}: 3,10,14,15$ $ F_{9}: 3,7,16$ $ F_{10}: 3,8,11,17$ $ F_{11}: 4,8,13,15$ $ F_{12}: 4,10,11,16$ $ F_{13}: 4,7,12,17$ $ F_{14}: 5,9,14,16$ $ F_{15}: 5,10,17$ $ F_{16}: 6,10,13$ $ F_{17}: 6,9,12,15$ Can somebody find something better?

Number1 wrote: Can somebody find something better? $ 71$ is best possible. Here's a configuration for 71: $ 1,2,3,4,5,6$ $ 1,7,11,15$ $ 1,8,12,16$ $ 1,9,13,17$ $ 2,10,14,16$ $ 2,8,13,15$ $ 3,9,14,15$ $ 3,10,11,17$ $ 4,8,14,17$ $ 4,9,11,16$ $ 4,10,12,15$ $ 5,7,14$ $ 5,15,16,17$ $ 6,7,8,9,10$ $ 6,11,12,13,14$ The missing pairs are $ (8,11),(9,12)$, and $ (10,13)$. A proof that we can't get $ 72$: Using Number1's notation, the only ways $ 72$ is possible are if $ f_{i}'=4$ for two $ i$, and $ f_{i}'=3$ for all other $ i$, or if $ f_{i}'=4$ for three $ i$, $ f_{i}'=2$ for one $ i$, and $ f_{i}'=3$ for all other $ i$. Suppose we have such a configuration. WLOG, order the fans so that the $ f_{i}'$ are in nonincreasing order. Case 1 ($ 4\cdot 2+3\cdot 14$): Now, for each $ j\in\{7,8,\cdots,17\}$, let $ a_{j}$ be the number of fans attending match $ j$. We have $ a_{j}\le 5$ for each $ j$, since each fan attending match $ j$ attends at least two other matches from the $ 11$, and these do not overlap. In addition, if $ a_{j}=5$, match $ j$ was not attended by fans $ 2$ or $ 3$; with ten matches to spread among five fans, we can't afford to put three matches with one fan. Since $ \sum_{j=7}^{17}a_{j}=50$, at least six of the $ a_{j}$ are equal to $ 5$. The two fans with $ f_{i}'=4$ cover at least seven matches between them, leaving only four other $ j$ that can possibly have $ a_{j}=5$. This is a contradiction, and this configuration is impossible. Case 2 ($ 4\cdot 3+3\cdot 12+2\cdot 1$): Define $ a_{j}$ as before. With only one fan that doesn't attend at least three of the $ 11$ matches, we still have $ a_{j}\le 5$ everywhere, and thus at least six of the $ a_{j}$ are equal to $ 5$. The only ways $ a_{j}$ can be $ 5$: either match $ j$ was attended by none of the fans $ 2,3,4$ or match $ j$ was attended by one of those fans and also by fan $ 17$. There are at most two matches in the first category (at least $ 4\cdot 3-3=9$ matches covered by $ 2,3,4$) and at most two in the second (everything fan $ 17$ attended), for a total of four. This is a contradiction, and this configuration is also impossible. [Edit- proof of maximality added]

I made a mistake in the example- here's a fixed version: $ 1,2,3,4,5,6$ $ 1,7,8,9,10$ $ 1,11,12,13,14$ $ 1,15,16,17$ $ 2,7,11,15$ $ 2,8,12,16$ $ 2,9,13,17$ $ 3,10,14,16$ $ 3,7,12,17$ $ 3,8,13,15$ $ 4,9,14,15$ $ 4,10,11,17$ $ 4,7,13,16$ $ 5,8,14,17$ $ 5,9,11,16$ $ 5,10,12,15$ $ 6,7,14$

jmerry wrote: Number1 wrote: Can somebody find something better? $ 71$ is best possible. Here's a configuration for 71: $ 1,2,3,4,5,6$ $ 1,7,11,15$ $ 1,8,12,16$ $ 1,9,13,17$ $ 2,10,14,16$ $ 2,8,13,15$ $ 3,9,14,15$ $ 3,10,11,17$ $ 4,8,14,17$ $ 4,9,11,16$ $ 4,10,12,15$ $ 5,7,14$ $ 5,15,16,17$ $ 6,7,8,9,10$ $ 6,11,12,13,14$ The missing pairs are $ (8,11),(9,12)$, and $ (10,13)$. A proof that we can't get $ 72$: Using Number1's notation, the only ways $ 72$ is possible are if $ f_{i}'=4$ for two $ i$, and $ f_{i}'=3$ for all other $ i$, or if $ f_{i}'=4$ for three $ i$, $ f_{i}'=2$ for one $ i$, and $ f_{i}'=3$ for all other $ i$. Suppose we have such a configuration. WLOG, order the fans so that the $ f_{i}'$ are in nonincreasing order. Case 1 ($ 4\cdot 2+3\cdot 14$): Now, for each $ j\in\{7,8,\cdots,17\}$, let $ a_{j}$ be the number of fans attending match $ j$. We have $ a_{j}\le 5$ for each $ j$, since each fan attending match $ j$ attends at least two other matches from the $ 11$, and these do not overlap. In addition, if $ a_{j}=5$, match $ j$ was not attended by fans $ 2$ or $ 3$; with ten matches to spread among five fans, we can't afford to put three matches with one fan. Since $ \sum_{j=7}^{17}a_{j}=50$, at least six of the $ a_{j}$ are equal to $ 5$. The two fans with $ f_{i}'=4$ cover at least seven matches between them, leaving only four other $ j$ that can possibly have $ a_{j}=5$. This is a contradiction, and this configuration is impossible. Case 2 ($ 4\cdot 3+3\cdot 12+2\cdot 1$): Define $ a_{j}$ as before. With only one fan that doesn't attend at least three of the $ 11$ matches, we still have $ a_{j}\le 5$ everywhere, and thus at least six of the $ a_{j}$ are equal to $ 5$. The only ways $ a_{j}$ can be $ 5$: either match $ j$ was attended by none of the fans $ 2,3,4$ or match $ j$ was attended by one of those fans and also by fan $ 17$. There are at most two matches in the first category (at least $ 4\cdot 3-3=9$ matches covered by $ 2,3,4$) and at most two in the second (everything fan $ 17$ attended), for a total of four. This is a contradiction, and this configuration is also impossible. [Edit- proof of maximality added] Why there cant be one 5 and rest 3-s?

Seventeen matches. If one match is attended by five and the rest by three each, that's a total of $5+3\cdot 16=53$ tickets. Possible, but nowhere close to the maximum we're looking for.

I had the same construction as jmerry but using different cases: Fill in the six games that person $17$ went to last. In total at most $22$ tickets can be bought for these games. If the most popular game was visited by $>6$ people then you can use simple bounding If the most popular game was visited by $6$ people then we get the construction above. If the most popular game was visited by $5$ people you get that at most $6$ games can be visited by five people by breaking up into the following cases for $3$ games visited by five people:

After working systematically with these cases you get at most $70<71$ tickets sold in total. So the answer is $71$. *This type of question was very popular from 2000-2003 similar to USAMO 2001 P1 and IMO 2001 P3. The only difference is that this question required a very long BASH.