If $n = 2j + 1$ is odd, Alina's strategy is the following: she cuts out a unit square and labels the columns from $-j$ to $j$, the column from which the first unit square has been removed receiving the label $0$. Starting at this point of the game, whenever Bogdan removes a square from column number $k \in \{-j, \dots, -2, -1, 1, 2, \dots, j\}$, Alina removes the unit square positioned in column number $-k$ and on the same row as the unit square removed by Bogdan in his last move. (If Bogdan removes the square remaining in column $0$ he loses instantly.) If on Bogdan's move the cylinder didn't lose its circular connection, it will not lose it after Alina's move either. Hence, Alina will never destroy the cylinder and, as the game is bound to finish sooner or later, Bogdan is the one who will lose the game. Alina wins.

If $n = 2j$ is even, Bogdan wins by adopting the following strategy: he labels the columns from $-j + 1$ to $j$, column $0$ being the one from which Alina has removed a square in her initial move. Bogdan removes a square from column $j$ (the one lying opposite to column $0$). From now on, if Alina removes a square from column number $k \in \{-j+1, \dots, -2, -1, 1, 2, \dots, j-1\}$, Bogdan removes the square situated in the same row, but in the opposite column, namely $-k$. (If Alina removes the remaining square from column $0$ or from column $j$, she loses instantly.) If Alina's move didn't dismantle the surface of the cylinder, Bogdan's move won't do it either. Eventually, Alina will dismantle the cylindrical surface and Bogdan will win.