1. Lemma Proof:
We need to prove that for all $i \in \{1, 2, \dots, 2^n - 1\}$, ${2^n \choose i}$ is even.

By Lucas' Theorem, for any non-negative integers $m$ and $n$ and a prime $p$, the binomial coefficient ${m \choose n}$ modulo $p$ can be determined by the base-$p$ representations of $m$ and $n$. Specifically,
$${m \choose n} \equiv \prod_{j=0}^k {m_j \choose n_j} \pmod{p}$$
where $m = m_k m_{k-1} \cdots m_0$ and $n = n_k n_{k-1} \cdots n_0$ are the base-$p$ representations of $m$ and $n$, respectively.

For $p = 2$, $m = 2^n$, and $i$, we have:
$${2^n \choose i} \equiv \prod_{j=0}^{n-1} {1 \choose i_j} \pmod{2}$$
where $i = i_{n-1} i_{n-2} \cdots i_0$ is the binary representation of $i$.

Since ${1 \choose 0} = 1$ and ${1 \choose 1} = 1$, but ${1 \choose i_j} = 0$ for $i_j > 1$, it follows that:
$${2^n \choose i} \equiv 0 \pmod{2}$$
for any $i$ that is not a power of 2. Therefore, ${2^n \choose i}$ is even for all $i \in \{1, 2, \dots, 2^n - 1\}$.

2. Application to the Problem:
We need to determine the number of good strings $S$ of length 99 with 66 $A$'s and 33 $B$'s such that for each $n$ (where $1 \leq n \leq 99$), the substring made from the first $n$ characters of $S$ has an odd number of distinct permutations.

The number of distinct permutations of a substring of length $n$ with $b$ $B$'s is given by ${n \choose b}$. For this to be odd, ${n \choose b}$ must be odd.

3. Analysis of Substrings:
- For $n = 64$, the substring must have 0 $B$'s because ${64 \choose x}$ is even for any $x \in \{1, 2, \dots, 33\}$. Thus, the first 64 characters must all be $A$'s.
- For $n = 65$, the substring must have 1 $B$ because ${65 \choose 1} = 65$ is odd.
- For $n = 66$, the substring must have 1 $B$ because ${66 \choose 1} = 66$ is even, so the 66th character must be $A$.
- For $n = 67$, the substring must have 2 $B$'s because ${67 \choose 2}$ is odd.
- For $n = 68$, the substring must have 2 $B$'s because ${68 \choose 2}$ is even, so the 68th character must be $A$.

Continuing this pattern, we find that the string must have the form:
$$A^{64} B A B A B \cdots B A B$$
where the first 64 characters are $A$'s, followed by alternating $B$'s and $A$'s.

4. Conclusion:
The only good string that satisfies the conditions is:
$$A^{64} B A B A B \cdots B A B$$

The final answer is $\boxed{ A^{64} B A B A B \cdots B A B }$