APMO

Let $n=p_1^{{\alpha }_1}\cdot \dots \cdot p_k^{{\alpha }_k}$ be the prime factorization of $n$ with $p_1<\dots <p_k$, so that $p(n)=p_k$ and $\sigma (n)=\left(1+p_1+\cdots +p_1^{{\alpha }_1}\right)\dots \left(1+p_k+\cdots +p_k^{{\alpha }_k}\right)$. Hence $p_k-1=\frac{\sigma (n)}{n}={\prod }_{i=1}^k\left(1+\frac{1}{p_i}+\cdots +\frac{1}{p_i^{{\alpha }_i}}\right)<{\prod }_{i=1}^k\frac{1}{1-\frac{1}{p_i}}={\prod }_{i=1}^k\left(1+\frac{1}{p_i-1}\right)\le {\prod }_{i=1}^k\left(1+\frac{1}{i}\right)=k+1$, that is, $p_k-1<k+1$, which is impossible for $k\ge 3$, because in this case $p_k-1\ge 2k-2\ge k+1$. Then $k\le 2$ and $p_k<k+2\le 4$, which implies $p_k\le 3$. 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 $$\{\begin{array}{l}{2}^{\alpha +1}-1=3^{\beta }\\ 3^{\beta +1}-1=2^{\alpha +2}\end{array}⟺\{\begin{array}{c}{2}^{\alpha +1}-1=3^{\beta }\\ 3\cdot (2^{\alpha +1}-1)-1=2\cdot 2^{\alpha +1}\end{array}⟺\{\begin{array}{r}{2}^{\alpha +1}=4\\ 3^{\beta }=3\end{array},$$ and $n=2^{\alpha }{3}^{\beta }=6$.