Solution:
The answer is $144$. Suppose that $n = p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}$ is the prime factorization of $n$, with $p_{1} < p_{2} < \cdots < p_{k}$ all distinct and $a_{i} > 0$ for $i = 1, \ldots, k$. A positive integer $d$ is a divisor of $n$ if and only if its prime factorization is of the form $d = p_{1}^{b_{1}} \cdots p_{k}^{b_{k}}$ with $0 \leq b_{1} \leq a_{1}, \ldots, 0 \leq b_{k} \leq a_{k}$ (the exponent $0$ is also allowed because not all prime factors of $n$ necessarily have to appear in the factorization of $d$ as well: for example, if all the $b_{i}$ are equal to $0$, one obtains the divisor $d = 1$). The possible choices of $d$ thus correspond to the possible choices of its factorization, namely $a_{1} + 1$ choices for the exponent $b_{1}$ (all the integers $0, 1, \ldots, a_{1}$), ..., $a_{k} + 1$ choices for the exponent $b_{k}$. Combining these choices in all possible ways, one obtains that the number $n$ has exactly $(a_{1} + 1) \cdot \cdots \cdot (a_{k} + 1)$ divisors.

To have exactly $15 = 3 \cdot 5$ divisors, there are thus only the following possibilities:

(i) $k = 1$, $a_{1} = 14$ hence $n = p_{1}^{14}$. The smallest positive integer of this type is evidently $2^{14} = 16384$.

(ii) $k = 2$, $a_{1} = 2$, $a_{2} = 4$ hence $n = p_{1}^{2} p_{2}^{4}$. The smallest positive integer of this type is evidently $2^{2} 3^{4} = 324$.

(iii) $k = 2$, $a_{1} = 4$, $a_{2} = 2$ hence $n = p_{1}^{4} p_{2}^{2}$. The smallest positive integer of this type is evidently $2^{4} 3^{2} = 144$.

Therefore the number sought is $144$.