1. **Understanding the Euler's Totient Function $\varphi(x)$**:
The Euler's Totient Function $\varphi(x)$ counts the number of integers from $1$ to $x$ that are coprime with $x$. For a prime number $p$, $\varphi(p) = p - 1$ because all numbers less than $p$ are coprime with $p$.

2. Initial Observations:
- If $1$ is among the $a_i$, then $\varphi(1) = 1$, which would mean $S$ contains duplicate elements, violating the distinctness condition.
- If $2$ is among the $a_i$, then $\varphi(2) = 1$ must be in $S$. Since $\varphi(3) = 2$, $3$ cannot be among the $a_i$ because there are no $a$ with $\varphi(a) = 3$. Thus, the only set $S$ with $2$ is $\{1, 2\}$.

3. **Considering $a_i > 2$**:
- For $a > 2$, $\varphi(a)$ is even. Therefore, $\varphi(a_1), \varphi(a_2), \ldots, \varphi(a_n)$ must all be even.
- This implies $a_1, a_2, \ldots, a_n$ are all odd because the integers in $S$ are consecutive.

4. Smallest Element Analysis:
- Let $a$ be the smallest of $a_1, a_2, \ldots, a_n$. Then $\varphi(a) - 1$ cannot be in the set since it would be odd and therefore one of the $a_i$, but $\varphi(a) - 1 < a$, a contradiction.
- Therefore, the smallest element of $S$ is $\varphi(a)$. We also realize that $\varphi(a) = a - 1$ because if there were a $b \neq a$ in $a_1, a_2, \ldots, a_n$ such that $\varphi(a) = b - 1$, we'd need $\varphi(b) < b$ to be in $S$, but the only element in $S$ smaller than $b$ is $\varphi(a)$, so we'd have $\varphi(a) = \varphi(b)$, which is a contradiction since $S$ must have $2n$ distinct elements.

5. Inductive Process:
- The two smallest elements of $S$ are $\varphi(a)$ and $a = \varphi(a) + 1$, in that order.
- Repeating the same process for $S - \{\varphi(a), a\}$, we find a similar conclusion.
- Proceeding inductively, it's possible to show that if $a_1 < a_2 < \ldots < a_n$, then the elements of $S$, in increasing order, are:
$$\varphi(a_1) < a_1 < \varphi(a_2) < a_2 < \ldots < \varphi(a_{n-1}) < a_{n-1} < \varphi(a_n) < a_n$$

6. Consecutive Numbers and Primes:
- Because they are all consecutive numbers, it means that $\varphi(a_i) = a_i - 1$ for all $1 \leq i \leq n$, which means that all $a_i$ are prime.
- Since they all form an arithmetic progression of ratio $2$, the largest possible value for $n$ is $n = 3$, achieved with:
$$\varphi(3) < 3 < \varphi(5) < 5 < \varphi(7) < 7$$
- This means that there are also examples for $n = 1$ and $n = 2$, namely, $\{\varphi(3), 3\}$ and $\{\varphi(3), 3, \varphi(5), 5\}$.

The final answer is $n = 1, 2, 3$