It is easy to compute $a_1=1$, $a_2=2$. Let us focus now on $a_{n+2}$ for $n\geq 1$. An $(n+2)$-level-good set $A$ not containing $n+2$ is also an $(n+1)$-level-good set (and vice-versa). An $(n+2)$-level-good set $A$ containing $n+2$ can be written as $A=A'\cup\{n+2\}$. If $A'\neq \emptyset$, then clearly $1< \min_{x\in A'} x \leq \max_{x\in A'} x \leq n+1$, and $A' - 1 = \{a - 1\mid a\in A'\}$ must be an $n$-level-good set; conversely, for $B$ being an $n$-level-good set, then $(B+1)\cup \{n+2\}$ is an $(n+2)$-level-good set. Counting the one case when $A'= \emptyset$ we get $a_{n+2} = a_{n+1} + a_n + 1$. Now, we can write that as $(a_{n+2} + 1) = (a_{n+1} + 1) + (a_n + 1)$, which allows us to affirm that $a_n = F_{n+2} - 1$, where $(F_n)_{n\geq 1}$ is the Fibonacci sequence $1,1,2,3,5,8,\ldots$.