Answer: $n=8,9,12$. Let $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ and $p_1<...<p_k$ since $n\neq 1$. $(\alpha_1+1)...(\alpha_k+1)=p_1^{\alpha_1-1}p_2^{\alpha_2}...p_k^{\alpha_k}$ Claim: $n$ has at most $2$ distinct prime divisors. Proof: Suppose that $k\geq 3$. Then we have $$\Pi{(\alpha_i+1)}=p_1^{\alpha_1-1}.p_2^{\alpha_2}...p_k^{\alpha_k}\geq 2^{\alpha_1-1}.3^{\alpha_2}.5^{\alpha_3+...+\alpha_k}$$$$1>2.\frac{2}{3}.\Pi{\frac{2}{5}}\geq (\frac{\alpha_1+1}{2^{\alpha_1-1}})(\frac{\alpha_2+1}{3^{\alpha_2}}).\Pi{(\frac{\alpha_i+1}{5^{\alpha_i}})}\geq 1$$Which is impossible.$\square$ If $n=p^{\alpha}.q^{\beta}$ and $p<q$, then $(\alpha+1)(\beta+1)=p^{\alpha-1}q^{\beta}$ $q^{\beta}\geq 3^{\beta}>\beta+1\implies \alpha+1>p^{\alpha-1}$ If $\alpha=1,$ then $2(\beta+1)=q^{\beta}\implies q=2$ which results in a contradiction. Hence $\alpha\geq 2$. Then $p=2$ and $\alpha=2$. We get $3(\beta+1)=2q^{\beta}\implies q=3\implies 2.3^{\beta-1}=\beta+1\implies \beta=1$ Thus we get $\boxed{n=12}$ If $n=p^\alpha,$ then $\alpha+1=p^{\alpha-1}$ Obviously $\alpha\neq 1$. If $p\geq 5,$ then $\alpha+1=p^{\alpha-1}\geq 5^{\alpha-1}\geq \alpha+3$ which is impossible. If $p=3,$ then $\alpha+1=3^{\alpha-1}\implies \alpha=2\implies \boxed{n=9}$ If $p=2,$ then $\alpha+1=2^{\alpha-1}\implies \alpha=3\implies \boxed{n=8}$