4. A positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that the sum of two of these divisors is six times the sum of the other two. Then $n=$ $\qquad$ .
4.287 or 143.

Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p<q$).
When $n=p^{3}$, all its positive divisors are $1, p, p^{2}, p^{3}$. According to the problem, we have
$$
\{i, j, k\}=\{1,2,3\},\{s, t\}=\{1,6\} \text {, }
$$

satisfying
$$
s\left(1+p^{i}\right)=t\left(p^{j}+p^{k}\right) .
$$

The right side of the above equation is a multiple of $p$, so $s=6, p=2$ or 3. Upon inspection, the equation does not hold in either case.
When $n=p q$, all its positive divisors are $1, p, q, p q$. Clearly, $p+q<p q$.
If $1+p q=6(p+q)$, then
$(p-6)(q-6)=35=1 \times 35=5 \times 7$
$\Rightarrow(p, q)=(7,41)$ or $(11,13)$
$\Rightarrow n=287$ or 143 .
If $p+p q=6(1+q)$, then $p=6$, which contradicts $p$ being a prime.

If $q+p q=6(1+p)$, then $q=6$, which contradicts $q$ being a prime.

In summary, $n=287$ or 143 .