Assume there exists a color function $f: Z \rightarrow\{R, B, G, Y\}$, such that for any integer $a$, we have
$$f\{a, a+x, a+y, a+x+y\}=\{R, B, G, Y\},$$

where $R$ represents red, $B$ represents blue, $G$ represents green, and $Y$ represents yellow. Let $g: Z \times Z \rightarrow\{R, B, G, Y\}$, and
$$g(i, j)=f(i x+j y) .$$

Thus, in the Cartesian coordinate system, each unit square's vertices have four different colors.
(1) If there exists a column of integer pairs $i \times Z$, such that $\left.g\right|_{i \times 2}$ is not a periodic function with period 2, then there exists a row of integer pairs $Z \times j$, such that $\left.g\right|_{Z \times j}$ is a periodic function with period 2.
Actually, if $\left.g\right|_{i \times z}$ is not a periodic function with period 2, then in this column, there must be three adjacent integers
$Y$
$R Y R$
with different colors, let's assume they are $B$. Considering the adjacent unit squares' vertices, we have $G B G R$, and thus
YRY
YRYRY
$B G B G B$ and so on. Therefore, we obtain three rows of integer pairs, such that the function $g$ restricted to these three rows is a periodic function with period 2.
(2) If for an integer $i, g_{i}=\left.g\right|_{z \times i}$ is a periodic function with period 2, then for every $j \in \mathbf{Z}$, $g_{i}=\left.g\right|_{z \times j}$ is a periodic function with period 2. If $i \equiv j(\bmod 2)$, then the range of $g$ is the same as the range of $g_{i}$; if $i \not \equiv j(\bmod 2)$, then the range of $g$ is the other two values different from the range of $g_{i}$.

Actually, for the integer points on the $i$-th row, let's assume they are $\cdots R B R B R B \cdots$, using the property of the unit square's vertices, we have $\cdots$. $\cdots$ RBRBR $\cdots$, and thus
$$\begin{array}{l} \cdots R B R B R \cdots \quad \cdots B R B R B \cdots \\ \cdots Y G Y G Y \cdots \text { or } \cdots \text { YGYGY } \cdots \\ \cdots R B R B R \cdots \quad \cdots R B R B R \cdots \end{array}$$

For the situation below the $i$-th row, we can reach the same conclusion.
By changing rows and columns, we can obtain the same conclusions as (1) and (2). Assuming the rows are periodic with period 2, and $g(0,0)=R$, $g(1,0)=B$, then $g(y, 0)=B$, where $y$ is an odd number. If $i$ is odd, then $g(Z \times\{x\})=\{Y, G\}$. Since $g(y, 0)=f(x, y)=g(0, x)$, this leads to a contradiction.

Let the subset with an odd number of elements be $X_{2}$, then the union of $X_{1}$ and $X_{2}$ is an odd subset. Conversely, any odd subset of $S_{n}$ can be written as the union of $X_{1}$ and $X_{2}$.
$X_{1}$ can be chosen in $2^{k}$ ways.
$X_{2}$ can be chosen in
$$\begin{aligned} & C_{l}^{1}+C_{i}^{3}+\cdots+C_{i}^{2 i-1}(2 i-1 \text { is the largest odd number not greater than } l \text {) } \\ = & \left.\frac{1}{2} \text { ( } C_{l}^{i}+C_{l}^{1}+\cdots+C_{l}^{l}\right) \\ = & 2^{\prime \cdot 1} \text { (ways). } \end{aligned}$$

Thus,
$$a_{n}=2^{k} \cdot 2^{l-1}=2^{n-1} .$$

By (1), we have
$$b_{n}=a_{n}=2^{n-1} \text {. }$$
(3) Let $A_{n}\left(B_{n}\right)$ represent the sum of the capacities of all odd (even) subsets of $S_{n}$.
If $n$ is odd $(n \geqslant 3)$.
All odd subsets of $S_{n}$ can be composed of the following two types of subsets: (1) odd subsets of $S_{n-1}$. (2) the union of each even subset of $S_{n-1}$ and the set $\{n\}$. Thus,
$$A_{n}=A_{n-1}+\left(B_{n-1}+n b_{n-1}\right)=A_{n-1}+B_{n-1}+n \cdot 2^{n-2} \text {. (by the conclusion of (1)) }$$

Similarly, we get
$$B_{n}=B_{n-1}+\left(A_{n, 1}+n \cdot a_{n-1}\right)=A_{n-1}+B_{n-1}+n \cdot 2^{n-2} .$$

Comparing equations (2) and (3), we get
$$A_{n}=B_{n} \text {. }$$

If $n$ is even $(n \geqslant 4)$.
All odd subsets of $S_{n}$ can be composed of the following two types of subsets: (1) all odd subsets of $S_{n-1}$; (2) the union of each odd subset of $S_{n-1}$ and the set $\{n\}$. Thus,
$$A_{n}=A_{n-1}+\left(A_{n-1}+n \cdot a_{n-1}\right)=2 A_{n-1}+n \cdot 2^{n-2} .$$

Similarly, we get
$$B_{n}=2 B_{n-1}+n \cdot 2^{n-2} \text {. }$$

By (4) and (5) and $A_{n-1}=B_{n-1}$, we get $A_{n}=B_{n}$.
In summary, we have proved that for any $n \geqslant 3, A_{n}=B_{n}$.
(4) The complement of $X$ in $S_{n}$ is denoted as $X$, then the sum of the capacities of $X$ and $X$ equals the capacity of $S_{n}$, i.e., $1+2+\cdots+n$ $=\frac{1}{2} n(n+1)$. Therefore, the sum of the capacities of all subsets of $S_{n}$ is
$$2^{n-1} \cdot \frac{1}{2} n(n+1)=2^{n-2} \cdot n(n+1) \text {. }$$

Since $A_{n}=B_{u}$, we have
$$A_{n}=\frac{1}{2} \cdot 2^{n-2} \cdot n(n+1)=2^{n-3} n(n+1), n \geqslant 3 .$$