Minimum number of generators for quotients of congruence subgroups of SL(2, Z) For a given positive integer $N$ let $L(N)$ denote the principal congruence subgroup of $\operatorname{SL}(2, \mathbb{Z})$ of level $N$. It is known that $L(N)$ is a finitely generated free group. Let $T(N)$ denote the minimum number of generators. Let $r$ be a number strictly less than $T(N)$.
Question: Is there a congruence subgroup $L$ of $L(N)$ such that the minimum number of generators of $L(N)/L$ is at least $r$?
 A: $\DeclareMathOperator\SL{SL}$Not always. For simplicity, take $N$ a large prime $p$. Following user44191's suggestion in the comments, take $r$ to be $T(N)-1$.
Since any congruence subgroup contains a principal congruence subgroup, we may as well assume $L$ contains a principle congruence subgroup. We can write this as $L(p^e M)$ where $M$ is relatively prime to $M$. Then
$$L(p)/ L(p^eM) = \SL_2( \mathbb Z/M) \times \ker ( \SL_2(\mathbb Z/p^e) \to \SL_2( \mathbb Z/p))$$
Now $\SL_2(\mathbb Z/M)$ is a quotient of $\SL_2(\mathbb Z)$ and so can be generated by $2$ elements. The kernel $\ker ( \SL_2(\mathbb Z/p^e) \to \SL_2( \mathbb Z/p))$ is a $p$-group, so it can be generated by a number of elements equal to its $p$-rank, which is $3$. Thus the product can be generated by $5$ elements. (In fact, if we're careful, I'm pretty sure we can reduce this down to $3$, but this is unnecessary.)
Taking $p$ large enough that the principle congruence subgroup $L(p)$ has more than $5$ generators (i.e. taking $p>3$), we get a counterexample.

In the comments you ask whether, for each congruence subgroup $L$, there are congruence subgroups $L'' \subseteq L' \subseteq L$ such that the number of generators of $L'/ L''$ is at least half the number of generators of $L$.
The answer is no.
To see this, choose our congruence subgroup $L$ to be $L(p)$ for $p$ a large prime as above, and $T(p)$ the number of generators of $L$.
Let $[L:L']$ be the index of $L'$ in $L$. Because $L(p)$ is a free group on $T(p)$ generators, its subgroup $L'$ is a free group on $(T(p)-1) [L: L']+ 1$ generators and thus has generator rank at least $(T(p)-1) [L: L']+ 1$.
On the other hand, because $L/ L''$ can be generated by $5$ elements, it admits a surjection from a free group on $5$ generators, so its subgroup $L'/L''$, of index $[L:L']$, admits a surjection from a free group on $(5-1)[L:L']+1$ generators and thus has generator rank at most $4 [L:L']+1$.
To produce a counterexample, it suffices to have
$$ 4 [L:L']+1 <  \frac{ (T(p)-1) [L: L']+ 1}{2} $$
for which, given $[L:L']\geq 1$, it suffices to have $T(p)>10$, something that is easy to achieve for $p$ large enough.
