Solution:

Suppose it is not possible. Take any person, label him $P_1$. Suppose he should be in seat $S_1$. If seat $S_1$ is vacant, then we can just move him to $S_1$, so $S_1$ must be occupied by someone. Call him $P_2$. Continue, so that we get a sequence $P_1$, $P_2$, $P_3$, ... where $P_i$ should be in the seat occupied by $P_{i + 1}$. Since there are only finitely many people, we must get a repetition. Suppose the first repetition is $P_i = P_{i + j}$. Then we can move $P_i$ to $P_{i + 1}$, $P_{i + 2}$ to $P_{i + 3}$, ..., $P_{i + j - 1}$ to $P_i$ and then these $j$ people will all be in their correct seats. Contradiction. So it is possible.

Suppose that $m + n - 1$ tickets to seat $(1,1)$ have been sold and $n - 1$ seats to each of $(2,1)$, $(3,1)$, ..., $(m,1)$. Then to comply with the conditions the people with tickets to $(1,1)$ must occupy the whole of the first row and first column. Hence those with tickets to $(k,1)$ for $k > 1$ must occupy the whole of row $k$ apart from $(k,1)$. Thus the seating is completely determined and only one person is in the correct seat - namely the person in $(1,1)$.