Let $a_{n}$ be the number of well-filled subsets whose maximum element is $n$ (setting $a_{0}=1$). Then it's easy to see that $a_{2k+1}=a_{2k}+a_{2k-1}+\cdots+a_{0}$ and $a_{2k+2}=(a_{2k+1}-C_{k})+a_{2k}+\cdots+a_{0}$ where $C_{k}$ is the number of well-filled subsets of size $k+1$ with maximal element $2k+1$. We proceed to compute $C_{k}$. One can think of such a subset as a sequence of numbers $1 \leq s_{1}<\cdots<s_{k+1} \leq 2k+1$ such that $s_{i} \geq 2i-1$ for every $1 \leq i \leq k+1$. Equivalently, letting $s_{i}=i+1+t_{i}$ it's the number of sequences $0 \leq t_{1} \leq \cdots \leq t_{k+1} \leq k+1$ such that $t_{i} \geq i$ for every $i$. This gives the list of $x$-coordinates of steps up in a Catalan path from $(0,0)$ to $(k+1, k+1)$, so $C_{k}=\frac{1}{k+2}\binom{2(k+1)}{(k+1)}$ is equal to the $(k+1)$th Catalan number. From this we can solve the above recursion to derive that $a_{n}=\binom{n}{\lfloor(n-1) / 2\rfloor}$. Consequently, for even $n$, $a_{0}+\cdots+a_{n}=a_{n+1}=\binom{n+1}{\lfloor n / 2\rfloor}$. Putting $n=42$ gives the answer, after subtracting off the empty set (counted in $a_{0}$).