One-dimensional packing problem: Optimal decomposition of music structure
If $n'$ is the number of groups, this problem admits no $2^{o(n')}$-time algorithm for any choice of a constant $\epsilon &gt; 0$, unless the exponential time hypothesis (ETH) fails.
Let $G = (V, E)$ be an undirected graph with $n$ vertices $v_1, \dots, v_n$ and $m$ edges $e_1, \dots, e_m$.
Construct an instance of your problem  by creating $n'=n$ groups as follows:

For each vertex $v_i \in V$, add a new group with fitness $1$ containing all intervals $[j-1, j]$ where $j$ is such that $v_i$ is an endpoint of $e_j$.

Any independent set $S \subseteq V$ of $G$ induces of collection of non-intersecting groups (except for the boundaries of the intervals) with total fitness $|S|$.
On the converse, any subset of $S'$ non-intersecting groups must have total fitness $|S'|$ and induces an independent set of size $|S'|$ in $G$.
The claim follows since, assuming the ETH, there exists no $2^{o(n)}$-time algorithm for independent set.
Moreover, assuming $\mathsf{P} \neq \mathsf{NP}$, the above reduction also shows that for any constant $\epsilon&gt;0$ there exists no polynomial-time $n^{1-\epsilon}$-approximation algorithm for your problem, even in the special case of unitary fitness. See, e.g., here.