1.1.3 $\star \star$ A 4-element real number set $S$ has the sum of the elements of all its subsets equal to 2008 (here the sum of elements of the empty set is considered to be 0). Find the sum of all elements of $S$.
Let $S=\{a, b, c, d\}$, then $(a+b+c+d) \times 2^{3}=2008$ (since each element in $S$ appears in exactly $2^{3}$ subsets of $S$), hence $a+b+c+d=251$. The answer we are looking for is 251.