We claim that all pairs $(a,b)$ such that $a+b\geq n+1$ work. Assume first $a+b\geq n+1$. Then simply take a set $K$ of $b$ keys and a set $L$ of $a$ locks and test each key in $K$ against each lock in $L$. The condition $a+b\geq n+1$ guarantees that there is at least one matching pair. Now assume $a+b\leq n$ and let $K$ be the set of keys and $L$ the set of edges. For each $k\in K,l\in L$ draw an edge $(k,l)$ if key $k$ is tested against lock $l$. This way, we obtain a bipartite graph. We will show that this graph's complement contains a perfect matching, which proves that there is a way to arrange the keys such that no key is checked against the correct lock. For this, we apply Hall's theorem. Let $L'\subseteq L$ be any subset of locks and $K'$ the corresponding set of keys which are not tried on any of the locks in $L'$. We need to show $|K'|\geq |L'|$. Assume $|K'|<|L'|$. By definition, all keys in $K\setminus K'$ are tried on all locks $L'$. Each such key is tested $|L'|$ times, which is only possible if $a\geq |L'|$. Similarly, we test each lock in $L'$ with $|K\setminus K'|=n-|K'|$ keys, so $b\geq n-|K'|$. Adding up the inequalities, we obtain $a+b\geq n+|L'|-|K'|\geq n$, a contradiction to our assumption. Hence, there is such a strategy if and only if $a+b\geq n+1$.