Solution:

We claim that any such integer $n$ must be either equal to $1$ or of the form $2^{a} 3^{b}$, where $a \geq 1$ and $b \geq 0$.

First, we note that if $n > 1$, it must be even. This is because if $n$ admits a prime factorization $n = \prod_{i=1}^{k} p_{i}^{r_{i}}$ over distinct primes $p_{i}$, then $\varphi(n) = \prod_{i=1}^{k} (p_{i} - 1) p_{i}^{r_{i} - 1}$. In particular, if an odd prime $p_{i}$ divides $n$ then $p_{i} - 1$ divides $\varphi(n)$, and consequently $\varphi(n)$ is even.

Then, we note that at most one odd prime divides $n$. Suppose otherwise, i.e., $2^{a} p_{1}^{r_{1}} p_{2}^{r_{2}}$ is a part of the prime factorization of $n$ for odd primes $p_{1}, p_{2}$. Then by the multiplicativity of $\varphi$, we know that
$$\varphi\left(2^{a} p_{1}^{r_{1}} p_{2}^{r_{2}}\right) = 2^{a-1} (p_{1} - 1) p_{1}^{r_{1} - 1} (p_{2} - 1) p_{2}^{r_{2} - 1}$$
must divide $\varphi(n)$. Note that the largest power of $2$ dividing $n$ is $2^{a}$, but since $p_{1} - 1$ and $p_{2} - 1$ are both even, $2^{(a-1)} 2^{2} = 2^{a+1}$ divides $\varphi(n)$. This is a contradiction.

Finally, we show that the said odd prime $p$ dividing $n$ must in fact be equal to $3$. Indeed, if $n = 2^{a} p^{r}$ where $a$ and $r$ are positive integers, we have
$$\frac{n}{\varphi(n)} = \frac{2p}{p-1} = 2 + \frac{2}{p-1}.$$
For this to be an integer, $p = 3$.

Hence, we count all such integers of the form $n = 2^{a} 3^{r}$ by summing over all possible values of $r$ (because there are fewer):

If $r = 0$, then $0 \leq a \leq 6$. (This is the only case in which $a = 0$ is allowed.)

If $r = 1$, then $1 \leq a \leq 5$.

If $r = 2$, then $1 \leq a \leq 3$.

If $r = 3$, then $a = 1$.

This gives us a total of $7 + 5 + 3 + 1 = 16$ values of $n$.