Let $p$ is minimal prime divisor of $n$. Then $n>p^2$, because $n$ had at least four divisors. Therefore $m=\frac{n}{p}-p$ is divisor of $n$ or $n=qm\to np=qn-p^2\to (q-p)n=p^2.$ It give contradition with $n>p^2$.

Rust, there is a mistake in your solution: $n=qm\implies np=qn-p^2q$. Furthermore, a contradiction is not what we want because $6$ is a solution, for example .

We distinguish numbers by their parity: If $n$ is odd, by (a) it has two divisors $1 < a < b < n$, but $b - a$ is even so this doesn't work. Otherwise, $n$ is even. As $2$ and $4$ don't work by (a), we have $\frac{n}{2}>2$. Let $a = 2$, $b = \frac{n}{2}$. From $3(\frac{n}{2}-2) = n + (\frac{n}{2}-6)$, we have $2 < \frac{n}{b-a} < 3$ if $\frac{n}{2} > 6$, which would contradict (b). So the solutions are $6, 8, 12$.

Hmm I don't quite understand what you're doing r3d30m3j... All I am sure about is that there is an error because you miss a solution...

After we rule out $n > 12$, we have to check all even $n \leq 12$, and it seems I forgot $n=8$ ... Also, one should read $b-a$ for $a-b$ in the above. I think it has been fixed now...

Suppose that $n$ is odd, and satisfies all the conditions. Then all its factors are odd. Now there exists factors $a$, $b$ of $n$ s.t. $1<a<b<n$ Then $b-a$ is even so it is not a factor of $n$, contradiction by using (b). So $n$ must be even, so let $n=2k$ $2$ and $4$ do not have four distinct factors so $n>4$ so $k>2$ $1<2<k<2k$ so $k-2|2k$ (by using (b)) But $k-2|2k-4$ so $k-2|(2k)-(2k-4)=4$ So $k-2=1\text{ or } 2\text{ or } 4$, since $k>2$ i.e. $k-2>0$ So $k=3\text{ or } 4\text{ or } 6$ So $n=6\text{ or } 8\text{ or } 12$ When $n=6$, $(a, b, b-a)=(2, 3, 1) $ and $b-a$ divides $n$. When $n=8$, $(a, b, b-a)=(2, 4, 2)$ and $b-a=2$ divides $n$. When $n=12$, $(a, b, b-a)=(2, 3, 1); (2, 4, 2); (3, 4, 1)$ and in each case $b-a$ divides $n$. Hence $n=6$ or $n=8$ or $n=12$.

r3d30m3j wrote: After we rule out $n > 12$, we have to check all even $n \leq 12$, and it seems I forgot $n=8$ ... Also, one should read $b-a$ for $a-b$ in the above. I think it has been fixed now... But still for $n=8$ with the divisors $4$ and $2$, your inequality does not hold...

Write $n=p_1^{a_1}...p_k^{a_k}$,with $p_1^{a_1}<...<p_k^{a_k}$. Suppose $k>1$.Then, $P=p_2^{a_2}...p_k^{a_k}-p_1^{a_1}| n$. But $P$ is coprime with all prime divisors of $n$,thus $P=1$. We can clearly see it is impossible for $k\geq 3$(because $P>1$). Then $k=2$. So we get $p_2^{a_2}-p_1^{a_1}=1$ which due to parity and a well known result gives the solutions $p_2=3,a_2=2,p_1=2,a_2=3$ and $p_2=3,a_2=1,p_1=2,a_2=1$. Now, suppose $k=1$. Then $n=p^a$(by conditions of problem $a \geq 3$) and $p^{a-1}-p=p(p^{a-2}-1)|n$ which is clearly impossible unless $p=2,a=3$. So our solutions are $n=6,8,18$.

if $n$ is odd, then all its divisors are also odd,let its two different divisors $a$ and $b$ be $a-b$ even, this is a contradiction because the number $n$ is divisible by $a-b$. $n=2t$ $t >2$ $1<2<t<2t$ $t-2|2t \implies t-2|4 \implies t=3,4,6$ Answer:$\boxed {6,8,12}$