Perfect matching problem
This problem is indeed $NP-complete$ as you have suspected. To see this, we will show a reduction from $SubsetSum$.
The reduction
Let $A,s$ be an instance of $SubsetSum$, where $|A|=n$. We will define $L$ as all the numbers in $A$, in addition to another $n$ zeros. For example, if $A= 4, 5, 8$ then $L=4,5,8,0,0,0$.
Now, we define $M$ to be with $2n$ values, such that we have $n$ ones and $n$ zeros.
The target value will be $s$
Proof
Such a perfect matching defines us a subset $A'\in A$, that $a\in A'$ iff $1$ is matched with $a$. Since we have $n$ values in $A$, and we have $n$ ones and $n$ zeros in $M$, then each pair of $(0,1)$ in $M$ can be assigned to every pair $(a, 0)$ where $a\in A$, and thus any $a\in A$ can be matched with either $0$ or $1$, contributing to the total sum either $a$ or $0$. Therefore, there is a perfect match if and only if there is a subset with that sum (as the value of the matching is a sum of the subset $A'$).