Perfect matchings in a chessboard?
The answer is NOT $\lceil \frac{mn}2\rceil$ for all large $m, n$ if e.g. $p=6$ and $q=3$. Why? Notice that because of the remainders $\mod 3$ now the graph is the (vertex) disjoint union of three bipartite graphs and from each we can select the bigger half. For example if $m=n=100$, then this way we can place (at least) 5002 knights. (This is because $x+y \mod 6$ has six classes which are in three pairs, the differences between the cardinalities of the pairs is $1,1,2$.)

I do not know what happens if we add the condition that $p$ and $q$ are relative primes. (Note that apart from the 2 divisor this is equivalent to $p+q$ and $p-q$ being relative primes, in fact this is the condition that we need and which also shows that $p+q$ being odd is necessary.)