Nonterminal descriptional complexity of regular languages
Lemma 1. Consider $L = (ab)^* + (ba)^*$:

There exists CFG with two variables which generates $L$
There exist no CFG with one variable which generates $L$

For proving (1) we may just consider the following CFG $$ S \rightarrow bAa \mid A , \, A \rightarrow abA \mid \varepsilon$$
The second proposition is a bit more tricky. Suppose that there exists CFG $G = (\{S\}, \{a, b\}, P, S)$ such that $L = L(G)$. Every production is $S \rightarrow \alpha$ for some $\alpha \in \{a, b, S\}^*$. Say, $\alpha = w_1 S w_2 S \ldots w_k S w_{k+1}$. Then $$u_1, \ldots u_k \in L \Rightarrow w_1 u_1 w_2 u_2 \ldots w_k u_k w_{k+1} \in L,$$ because as $L$ is generated by $G$, one can derive this word by using $S \rightarrow w_1 S w_2 S \ldots w_k S w_{k+1}$ and then deriving every $u_i$ from $S$'s. By using that we can prove that in right-hand side of every production $S$ can't be followed immediately by any symbol:

if $S$ is followed by $a$, we have subword $Sa$ in $\alpha$, then we can derive $ba$ from $S$, thus we can derive the word which contains two $a$'s in the row, which obviously doesn't lie in $L$;
if $S$ is followed by $b$, having subsword $Sb$ of $\alpha$ we can derive $ab$ from $S$ and thus obtain the word which contains two $b$'s in the row;
if we have subword $SS$ of $\alpha$, we can derive $ab$ from the first $S$ and $ba$ from the second.

Also $S$ can't follow neither letter nor $S$ in right-hand side of every production for the same reason. So $G$ can only have productions of sort $S \rightarrow S$ and $S \rightarrow w$ for $w \in L$. Thus $L(G)$ is finite, which is contradiction. $\Box$
I'd also like to suggest the following
Lemma 2. Let $w_1, \ldots w_n$ be some words. The language $w_1^* + \ldots + w_n^*$ is generated by CFG with one variable iff words $w_1, \ldots w_n$ commute pairwise, i. e. $w_i w_j = w_j w_i$ for every $i,j$.