Three, let $S_{\mathrm{n}}$ denote the sum of the elements of all subsets of the set of natural numbers $\{1,2,3, \cdots, n\}$ (the sum of elements of the empty set is defined as 0). Find $S_{1980}$.
Three, Solution: Divide all subsets of the set $\{1,2,3, \cdots, n\}$ (a total of $2^{\text {D}}$ subsets) into two categories:
1. All subsets that do not contain the element $n$ have $2^{\mathrm{n}-1}$ subsets.
2. All subsets that contain the element $n$ also have $2^{\mathrm{n}-1}$ subsets.
Establish a one-to-one correspondence between these two categories of subsets: Let the universal set $I=\{1,2,3, \cdots, n\}$, for a subset $A$ in the first category, its complement $\bar{A}$ must belong to the second category (since $A$ does not contain the element $n$, then $\bar{A}$ must contain $n$), and different subsets in the first category must have different complements in the second category. Let the subset $A$ in the first category correspond to $\bar{A}$ in the second category, which is clearly a one-to-one correspondence.
$$
\because A+\bar{A}=I=\{1,2,3, \cdots, n\} \text {, }
$$
$\therefore$ The sum of the elements of each pair of corresponding subsets is
$$
1+2+3+\cdots+n=\frac{n(n+1)}{2} \text {. }
$$

Since there are $2^{\mathrm{D}-1}$ pairs, the required sum is
$$
S_{\mathrm{n}}=2^{\mathrm{a}-1} \cdot \frac{n(n+1)}{2}=2^{\mathrm{n}-2} \cdot n(n+1) \text {. }
$$

Therefore, $S_{1080}=2^{1081} \cdot 1989 \cdot 1990$.