For positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including $1$ and $n$ itself).

Firstly, note that $1$ and $2$ are good, since $1 = \frac{2}{d(2)}$ and
$$2 = \frac{8}{d(8)}.$$

Secondly, we note that if $p$ is an odd prime, then $p$ is good. This is because $d(8p) = 8$. In particular, $3, 5, 7, 11, 13, 17$ are good numbers.

Thirdly, we note that if $p$ is an odd prime, then $2p$ is good. This is because $d(2^2 \cdot 3^2 p) = 3 \cdot 3 \cdot 2$. In particular, $6, 10, 14$ are good numbers.

Fourthly, we note that
$$4 = \frac{36}{d(36)}, \quad 8 = \frac{96}{d(96)}, \quad 9 = \frac{108}{d(108)},$$
$$12 = \frac{240}{d(240)}, \quad 15 = \frac{360}{d(360)}, \quad 16 = \frac{128}{d(128)}.$$
Thus, the numbers $1, 2, \ldots, 17$ are good.

Finally, we prove that $18$ is not good. We approach indirectly by assuming that $18 = \frac{n}{d(n)}$ or $n = 18 d(n)$ for $n = 2^a \cdot 3^{b+1} \cdot p_1^{k_1} \cdots p_m^{k_m}$ (where $p_1 < \cdots < p_m$ are prime numbers greater than $3$ and $a, b, k_1, \cdots, k_m$ are positive integers); that is,
$$2^{a-1} \cdot 3^{b-1} \cdot p_1^{k_1} \cdots p_m^{k_m} = (a+1)(b+2)(k_1+1)\cdots(k_m+1). \qquad \textcircled{1}$$
For every odd prime $p$ and every positive integer $k$, we can show (by an easy induction on $k$) that
$$p^k > k+1. \qquad \textcircled{2}$$
Combining the last two relations, we deduce that
$$2^{a-1} \cdot 3^{b-1} < (a+1)(b+2)$$
or
$$f(a) = \frac{2^{a-1}}{a+1} < \frac{b+2}{3^{b-1}} = g(b).$$
It is easy to prove that $f(1) = \frac{1}{2}$, $f(2) = \frac{2}{3}$, $f(3) = 1$, $f(4) = \frac{8}{5}$, $f(5) = \frac{16}{6}$, and $f(a) \ge \frac{32}{7} > 4$ for $a \ge 6$. It is also easy to prove that $g(1) = 3$, $g(2) = \frac{4}{3}$, $g(3) < \frac{5}{9}$, and $g(b) < \frac{2}{9}$ for $b \ge 4$. Thus $\textcircled{1}$ holds only if $b \le 3$.

If $b=3$, then $(a, b) = (1, 3)$, and $\textcircled{1}$ becomes
$$9p_1^{k_1} \cdots p_m^{k_m} = 10(k_1 + 1)\cdots(k_m + 1),$$
implying that $p_1 = 5$. Since $\textcircled{2}$ and $\frac{p_1^{k_1}}{k_1+1} = \frac{5^{k_1}}{k_1+1} \ge \frac{5}{2}$ for positive integer $k_1 \ge 1$, we can easily see that there is no solution in this case.

If $b=2$, then $(a, b) = (1, 2)$, $(2, 2)$, $(3, 2)$, and $\textcircled{1}$ becomes
$$3 \cdot 2^{a-1} p_1^{k_1} \cdots p_m^{k_m} = 4(a+1)(k_1+1) \cdots (k_m+1).$$
We deduce that $4$ divides $2^{a-1}$ or $a \ge 3$. Hence $a$ must be equal to $3$. But then $4(a+1) = 16$ divides $2^{a-1}$, which is impossible.

If $b=1$, then $(a, b) = (1, 1)$, $(2, 1)$, $(3, 1)$, $(4, 1)$, $(5, 1)$, and $\textcircled{1}$ becomes
$$2^{a-1} p_1^{k_1} \cdots p_m^{k_m} = 3(a+1)(k_1+1) \cdots (k_m+1),$$
which is impossible since $p_i$ are primes greater than $3$.

In all the cases, we cannot find $n$ satisfying the condition $18 = \frac{n}{d(n)}$; that is, $18$ is not a good number.