Consider a graph with $50$ vertices, the cities representing the vertices, and edges representing the flights. Call a graph good if there are antitours from each city. Claim: A good graph cannot have any odd cycle. Proof: We assume the contrary. Let the odd cycle be $v_1v_2v_3 \dots v_{2k+1}$ Now consider the antitour from $v_1$. Let the antitour be $\{ C_1, C_2, C_3, \dots , C_{50} \}$. Now, as there are $2k+1$ vertices in the odd cycle, atleast one of them have index $\ge 2k+1$ in the antitour , i.e, $v_t=C_n$, ($n \ge 2k+1$), $v_t \in$ odd cycle. If $t,n$ are both odd, consider the path $\{ \underbrace{v_1v_{2k+1}v_{2k} \dots v_tv_{t-1}v_tv_{t-1} \dots}_{n \text{ terms}} \}$ If $t$ even, $n$ odd, $\{ \underbrace{v_1v_2v_3 \dots v_tv_{t+1}v_tv_{t+1} \dots}_{n \text{ terms}} \}$ If $t$ odd, $n$ even, $\{ \underbrace{v_1v_{2k+1}v_{2k} \dots v_tv_{t-1}v_tv_{t-1} \dots}_{n \text{ terms}} \}$ If $t, n$ are both even, $\{ \underbrace{v_1v_{2k+1}v_{2k} \dots v_tv_{t-1}v_tv_{t-1} \dots}_{n \text{ terms}} \}$ Above 4 paths contradicts that $v_t=C_n$, and we are done. Now let the capital city be $v_1$. As given, there is a path from $v_1$ to $v_{50}$ (WLOG) in the graph. As there are no odd cycles, there is no edge $\{v_iv_j\}$ if $i$ and $j$ have same parity. Claim: The minimum number of antitours from the capital city is $(25!)^2$ Proof: As there is no edge $\{v_iv_j\}$ for same parity $i, j$, each edge takes us from an odd index to an even index vertex. Now assign a random odd number to all $v_i$ if $i$ is odd, and even number to $v_i$ if $i$ is even and form the collection, $\{C_1, C_2, \dots, C_{50} \}$ This is clearly an antitour from $v_1$ as an odd number of edges taken from $v_1$ take us to an even numbered vertex, and vice versa. The number of ways to assign odd numbers is $25!$, and number of ways to assign even numbers is $25!$. Hence, there are atleast $(25!)^2$ possible antitours from $v_1$. For the construction, join $v_1$ to all vertices $v_i$ if $i$ is even. and form the $v_1v_2 \dots v_{50}$ cycle. Clearly, this works.

are there official sols??