Asian Pacific Mathematics Olympiad

Let $n=p_{1}^{\alpha_{1}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}$ be the prime factorization of $n$ with $p_{1}<\ldots<p_{k}$, so that $p(n)=p_{k}$ and $\sigma(n)=\left(1+p_{1}+\cdots+p_{1}^{\alpha_{1}}\right) \ldots\left(1+p_{k}+\cdots+p_{k}^{\alpha_{k}}\right)$. Hence
$p_{k}-1=\frac{\sigma(n)}{n}=\prod_{i=1}^{k}\left(1+\frac{1}{p_{i}}+\cdots+\frac{1}{p_{i}^{\alpha_{i}}}\right)<\prod_{i=1}^{k} \frac{1}{1-\frac{1}{p_{i}}}=\prod_{i=1}^{k}\left(1+\frac{1}{p_{i}-1}\right) \leq \prod_{i=1}^{k}\left(1+\frac{1}{i}\right)=k+1$,
that is, $p_{k}-1<k+1$, which is impossible for $k \geq 3$, because in this case $p_{k}-1 \geq 2 k-2 \geq k+1$. Then $k \leq 2$ and $p_{k}<k+2 \leq 4$, which implies $p_{k} \leq 3$.
If $k=1$ then $n=p^{\alpha}$ and $\sigma(n)=1+p+\cdots+p^{\alpha}$, and in this case $n \nmid \sigma(n)$, which is not possible. Thus $k=2$, and $n=2^{\alpha} 3^{\beta}$ with $\alpha, \beta>0$. If $\alpha>1$ or $\beta>1$,
$$\frac{\sigma(n)}{n}>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)=2 .$$
Therefore $\alpha=\beta=1$ and the only answer is $n=6$.

Comment: There are other ways to deal with the case $n=2^{\alpha} 3^{\beta}$. For instance, we have $2^{\alpha+2} 3^{\beta}=\left(2^{\alpha+1}-1\right)\left(3^{\beta+1}-1\right)$. Since $2^{\alpha+1}-1$ is not divisible by $2$, and $3^{\beta+1}-1$ is not divisible by $3$, we have
$$\left\{\begin{array} { l } { 2 ^ { \alpha + 1 } - 1 = 3 ^ { \beta } } \\ { 3 ^ { \beta + 1 } - 1 = 2 ^ { \alpha + 2 } } \end{array} \Longleftrightarrow \left\{\begin{array} { c } { 2 ^ { \alpha + 1 } - 1 = 3 ^ { \beta } } \\ { 3 \cdot ( 2 ^ { \alpha + 1 } - 1 ) - 1 = 2 \cdot 2 ^ { \alpha + 1 } } \end{array} \Longleftrightarrow \left\{\begin{array}{r} 2^{\alpha+1}=4 \\ 3^{\beta}=3 \end{array},\right.\right.\right.$$
and $n=2^{\alpha} 3^{\beta}=6$.