Claim 1. Mona cannot prevent Bengt if she has at most $n-2$ prohibitions of each kind. Proof. Assume the contrary. Colour the $n \times n$-board with colours $1,2, \ldots, n$ such that square $(i,j)$, where $1 \leq i \leq n$ and $1 \leq j \leq n$, has the same colour as the remainder when $i+j-1$ is divided by $n$. Note that by construction, each row and column contain exactly one of every colour and the board contains exactly $n$ squares of each colour. For brevity, call such a set of $n$ squares good if all the squares have the colour. Since Mona is at disposal at most $n-2$ prohibitions of each kind, then she can block at most $n-2$ different good sets with rings and at most $n-2$ different good sets with crosses, thus leaving at least two goods sets $A$ and $B$ such that $A$ is free from cross-prohibitions and $B$ is free from ring-prohibitions. In other words, Bengt can place crosses in $A$ and rings in $B$, which means Mona has failed her goal. The claim is proven. This means that Mona has to use at least $n-1$ prohibitions of at least one of the two kinds, ie. we have achieved a lower bound of $n-1$. Claim 2. Using $n-1$ of each prohibitions is sufficient. Proof. An example (among many) which supports this claim is when Mona places $n-1$ prohibitions against rings in the topmost row, ignoring the first square, and $n-1$ prohibitions against crosses in the leftmost column, ignoring the first square. Hence the minimum is $n-1$.