If I understand the way things are counted, the maximum possible value of $\mathrm{al}(M)$ is $\binom{n/2}{2}$. Since this is less than half $\binom{n}{2}$, there is no way to put a matching having no aligned pairs with two other matchings so that the total number of alignments is $\binom{n}{2}$,

So the answer is "no".

If I understand the way things are counted, the maximum possible value of $\mathrm{al}(M)$ is $\binom{n/2}{2}$. Since this is less than half $\binom{n}{2}$, there is no way to put a matching having no aligned pairs with two other matchings so that the total number of alignments is $\binom{n}{2}$,

So the answer is "no".

ADDED: Withdrawn for the reasons noted in the comments.