4. Let the set of numbers $M=\{a, b, c, d\}$, and the pairwise sums of $a, b, c, d$ form the set $S=\{5,8,9,11,12,15\}$. Then $M=$ $\qquad$
4. $\{1,4,7,8\}$ or $\{2,3,6,9\}$.

Let $a<b<c<d$. Since set $S$ has $6=\mathrm{C}_{4}^{2}$ elements, it is known that the sums of any two of $a, b, c, d$ are all different.

Since $a+b<a+c<a+d<b+d<c+d$, and $a+c<b+c<b+d$,
there are only two cases:
(1) $a+d<b+c$, then
$$
\begin{array}{l}
(a+b, a+c, a+d, b+c, b+d, c+d) \\
=(5,8,9,11,12,15) .
\end{array}
$$

Thus, $c-b=3, b+c=11$.
Solving gives $b=4, c=7$.
Then, $a=1, d=8$, so $\{a, b, c, d\}=\{1,4,7,8\}$.
(2) $b+c<a+d$, then
$$
\begin{array}{l}
(a+b, a+c, b+c, a+d, b+d, c+d) \\
=(5,8,9,11,12,15):
\end{array}
$$

Thus, $c-b=3, b+c=9$.
Solving gives $b=3, c=6$.
Then, $a=2, d=9$, so $\{a, b, c, d\}=\{2,3,6,9\}$.