I claim $n=4$ or $n=p$, where $p$ is an arbitrary prime are the only answers. Clearly $n=4$ works, whereas the statement is vacuous as $n=p$. Now let $d(n)$ denotes the number of positive integer divisors of $n$. We inspect $d(n)\in\{3,4\}$. If $d(n)=3$ then $n=p^2$ for some prime $p$. As $p-1\mid p^2$ we get $p-1\mid 1$, forcing $p=2$ and $n=4$. Assume $d(n)=4$. Then either $n=p^3$ for a prime $p$ (which is impossible, as otherwise $p-1\mod p^3$ yields $p=2$, so $n=8$, but $4-1\nmid 8$), or $n=pq$ for primes $2\le p<q\le n$. But then $p-1\mid pq$ and $q-1\mid pq$. The former implies $p-1\mid q$ and the latter gives $q-1\mid p$; this is a contradiction as $p\ge q-1$ and $q\ge p-1$ cannot simultaneously hold.

grupyorum wrote: The former implies $p-1\mid q$ and the latter gives $q-1\mid p$; this is a contradiction as $p\ge q-1$ and $q\ge p-1$ cannot simultaneously hold. Actually, this can happen in a special case when $p=q-1$ so $p=2, q=3$. This gives us solution $n=6$ as well, otherwise your solution is completely fine.

If $d(n)=4$, it must be in the form $pq$ or $p^3$ where $p$ and $q$ are primes. If $n=p^3$, then $p^2-1=(p+1)(p-1)\nmid{n}$. If $n=pq$, WLOG let $q>p$, then $p-1\mid{p} \Longrightarrow p=2$. Then $q\neq 2$ so $\gcd{(q, q-1)}=1 \Longrightarrow q-1=p$. Then $pq=6$. If $d(n)=3$, it must be in the form $p^2$, so $p-1\mid{p} \Longrightarrow p=2$. Then, $p^2=4$. If $d(n)=2$, it is prime so it works.

I've been told by a Romanian friend that your translation of the problem is wrong. It originally asked for those $n\ge 2$ with at least 4 natural divisors. Also, less important, but since I'm pointing out inaccuracies, the original problem doesn't compare $d_1$ and $d_2$, so the word positive in "positive integer $d_1-d_2$" is absent in its statement.