, 2016

Solution:
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
$$\begin{aligned} & a_{2k+1} = a_{2k} + a_{2k-1} + \cdots + a_{0} \\ & a_{2k+2} = \left(a_{2k+1} - C_{k}\right) + a_{2k} + \cdots + a_{0} \end{aligned}$$
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 $a_{0}$).