A group $\mathfrak{G}$ with a maximal normal subgroup $N$ and $[\mathfrak{G}:N]=n$ exists if and only if there is exists a simple group of order $n$.
Indeed, if such a group exists, then $|\mathfrak{G}/N|=n$, and the maximality of $N$ means, by the correspondence theorem, that $G/N$ has no proper nontrivial normal subgroups; i.e., $G/N$ is simple of order $n$.
For the converse, while one can just take a nonabelian group $G$ and then consider $G\times S$, let’s look for a slightly more interesting example, one in which $S$ is not (usually) also normal.
to that end, let $S$ be as simple group of order $n$; let $G$ be any nontrivial group. We construct a group $\mathfrak{G}$ with a normal subgroup $N$ such that $\mathfrak{G}/N\cong S$. $\mathfrak{G}$ will be nonabelian.
The group is the standard (unrestricted) wreath product of $G$ by $S$, $\mathfrak{G}=G\wr S$, constructed as follows:
Let $B=G^S$, the set of all set theoretic functions from $S$ to $G$, endowed with the pointwise product. This is isomorphic to the direct product of $|S|$ copies of $G$, indexed by $S$.
Let $S$ act on $B$ on the right by letting $S$ act on the indices via the regular right action. That is, given $\mathbf{g}=(g_s)_{s\in S}\in B$, we let
$$\mathbf{g}\cdot t = (g_{st})_{s\in S}.$$
This action allows the construction of the semidirect product $B\rtimes S$. The standard unrestricted wreath product is given by that semidirect product, $G\wr S=B\rtimes S$.
This group is nonabelian whenever $G$ and $S$ are nontrivial. In addition, a theorem of Kaloujnine and Krasner shows that any group that is an extension of $G$ by $S$ can be realized as a subgroup of $G\wr S$.
The subgroup $B$ is normal in $G\wr S$ with $(G\wr S)/B\cong S$. Since $S$ was chosen to be simple, the correspondence theorem guarantees that $B$ is a maximal normal subgroup of $G\wr S$. The index of $B$ is $|S|$.
