Answer:

a) The maximal possible value of area equals $\frac{1}{3}(4^n - 1)(4^n + n - 1)$ and it is attained at sets $(1, 1, \dots, 1, 2^{2n})$ and $(2^2, 2^4, \dots, 2^{2n-2}, 1)$.

b) The minimal possible value equals $4(2^n - 1)^2$ and it is attained in two cases: either the sets $x_i$ and $y_i$ equal $(2^2, 2^3, \dots, 2^{n+1})$ and $(2^0, 2^1, \dots, 2^{n-1})$ or both sets equal $(2^1, 2^2, \dots, 2^n)$.

Since the lengths $x_i$ and $y_j$ are divisors of the given numbers, the total amount of sets is finite, so there exist the minimal area of the rectangle $OABC$. We will prove the following

Lemma. The area of the rectangle $OABC$ is maximal if and only if the next two conditions hold:

1) For each $i$ from $1$ to $n$ either $x_i$ or $y_i$ equals $1$;

2) The difference $|(x_1 + x_2 + \dots + x_n) - (y_1 + y_2 + \dots + y_n)|$ is minimal among all sets, satisfying condition 1).

Proof of the lemma. Take sets $(x_1, x_2, \dots, x_n)$ and $(y_1, y_2, \dots, y_n)$ such that the area of the rectangle $OABC$ is maximal.

First we prove the necessity of the first condition. Suppose there exists $k$ such that both $x_k$ and $y_k$ are greater than $1$. Denote $x_1 + x_2 + \dots + x_n = X$, $X - x_k = X_k$, $y_1 + y_2 + \dots + y_n = Y$ and $Y - y_k = Y_k$. Then the area of the rectangle $OABC$ equals $S = XY = (X_k + x_k)(Y_k + y_k)$. Without loss of generality let $X_k \ge Y_k$. Replace in the sets $(x_i)$ and $(y_i)$ the lengths $x_k$ and $y_k$ by $1$ and $S_k$, respectively. The area of the new rectangle equals $S' = (X_k + 1)(Y_k + S_k)$. The difference between new and old areas equals

$$S' - S = (X_k + 1)(Y_k + S_k) - (X_k + x_k)(Y_k + y_k) = (X_k y_k - Y_k)(x_k - 1) > 0,$$

which contradicts the minimality of $S$.

Now we prove the necessity of the second condition. The first condition implies

$$X + Y = S_1 + S_2 + \dots + S_n + n.$$

This sum depends only on the given values $S_1, S_2, \dots, S_n$, denote this sum by $C$. The area of the rectangle $OABC$ equals $S = XY = X(C - X)$, which is the quadratic polynomial of the variable $X$. The corresponding parabola is opened to the bottom and the abscissa of its vertex equals $C/2$ (which corresponds $X = Y$). Hence the area of the rectangle is maximal when $X$ is closest to $C/2$, which is equivalent to the minimality of $|X - Y|$.

To prove the sufficiency of the conditions 1) and 2) note that for all sets satisfying these conditions the values of $X$ and $Y$ are the same (up to permutation), hence they all provide the maximal value of the area. The lemma is proved.

Proceed to the solution of a). Since $2^{2n} = 1 + 2^0 + 2^1 + \dots + 2^{2n-1}$, we have

$$2^{2n} > 2^2 + 2^4 + \dots + 2^{2n-2},$$

whence the conditions of the lemma are satisfied if and only if one set equals $(1, 1, \dots, 1, 2^{2n})$ while the second set equals $(2^2, 2^4, \dots, 2^{2(n-1)}, 1)$. Wherein the maximal possible area of the rectangle equals $\frac{1}{3}(4^n - 1)(4^n + n - 1)$. Thus, a) is finished.

To find the minimal possible area, we use the Cauchy-Bunyakovsky inequality

$$(x_1 + x_2 + \dots + x_n)(y_1 + y_2 + \dots + y_n) \ge (\sqrt{x_1y_1} + \sqrt{x_2y_2} + \dots + \sqrt{x_ny_n})^2 = (\sqrt{S_1} + \sqrt{S_2} + \dots + \sqrt{S_n})^2.$$

If there exist at least one pair of sets $(x_i)$ and $(y_i)$ at which this inequality is the equality, then the last expression provides the minimal possible area. Moreover, then this minimal value is attained only for such sets $(x_i)$ and $(y_i)$. The Cauchy-Bunyakovsky inequality turns into the equality if and only if $\frac{x_1}{y_1} = \frac{x_2}{y_2} = \dots = \frac{x_n}{y_n}$. Since $x_1y_1 = S_1 = 4$, there exist the three variants: $x_1 = 4$, $y_1 = 1$; $x_1 = 1$, $y_1 = 4$; and $x_1 = y_1 = 2$. First two variants give the pair $(2^2, 2^3, \dots, 2^{n+1})$ and $(2^0, 2^1, \dots, 2^{n-1})$ of sets, and the third variant gives two sets equal $(2^1, 2^2, \dots, 2^n)$. Both cases provide the minimal possible area $4(2^n - 1)^2$.