, 2014

We show that this is the case if and only if $n$ is square-free, i.e. if $n$ contains no prime factor in a power greater than $1$. In order to see this, write
$$n = \prod_{j=1}^{r} p_j^{e_j}, \quad \text{and therefore} \quad d(n) = \prod_{j=1}^{r} (e_j + 1).$$
If all powers $e_j$ are equal to $1$, we have $d(n) = 2^r$. In this case, every divisor $t$ of $n$ can also be written as the product of $s$ distinct primes (with $0 \le s \le r$), and therefore $d(t) = 2^s$, which is certainly a divisor of $d(n) = 2^r$.

We now assume that at least one value of $e_j$ is greater than one, without loss of generality, let this be $e_r$. We now consider the divisor $t = \frac{n}{p_r} = p_r^{e_r-1} \cdot \prod_{j=1}^{r-1}(e_j+1)$. For this divisor, we have $d(t) = e_r \cdot \prod_{j=1}^{r-1}(e_j+1)$. In this case, we have $\frac{d(n)}{d(t)} = \frac{e_r+1}{e_r} = 1 + \frac{1}{e_r}$, which is certainly not an integer, and our proof is complete. $\square$