By the induction method we will show that divide in to two subset $A, B$ such that $a_i + b_i$ is prime number for $i|1, n$.
If $n = 1$ then $\{1, 2\} = \{1\} \cup \{2\}$, $1 + 2 = 3$ is prime number and $\{1\} \cap \{2\} = \emptyset$. Assume now that the result is proved for $1, 2, 3, ..., n - 1$. Now we will show that for $n$.
By well known Bertrand's postulate there exists prime number $p$ such that $2n < p < 4n$. Because of $p > 2n \ge 2$ then $p$-odd prime number. Thus $p = 2n + m$, $m$-odd positive integer. Now consider
$$(2n, m); (2n - 1, m + 1); \dots \left( \frac{2n + m + 1}{2}, \frac{2n + m - 1}{2} \right)$$
pairs and those pair's two numbers sum is $p = 2n + m$. For other remaining numbers are $1, 2, ..., m - 1$ get new set that is $\{1, 2, ..., m - 1\}$, which is by the induction method dividing into two subsets. Because $m - 1$ is even number. Thus the set $\{1, 2, 3, ..., 2n\}$ dividing into two subsets.

Let $(a_1 + b_1)...(a_n + b_n) = p_1^{\alpha_1} p_2^{\alpha_2}...p_s^{\alpha_s}$, $\alpha_1 + ... + \alpha_s = n$, $\alpha_i \in \mathbb{N}$.
Hence the number of divisors of $(a_1 + b_1)...(a_n + b_n)$ is $N := (\alpha_1 + 1)...(\alpha_s + 1)$.

If $\alpha_j = 1$ for some $j$ then
$$N = 2 \cdot (\alpha_1 + 1)(\alpha_2 + 1)...(\alpha_{j-1} + 1)(\alpha_{j+1} + 1)...(\alpha_s + 1).$$
So we need to prove that following inequality
$$\frac{\prod (\alpha_j + 1)}{\alpha_j + 1} \le 2^{n-1}.$$

This inequality is same as property (iii). Hence we can assume that $\alpha_j + 1$ for arbitrary $i \in \{1, 2, ..., s\}$. Using the Cauchy's inequality, we get that
$$\prod_{i=1}^{s} (\alpha_i + 1) \le \left( \frac{\sum_{i=1}^{s} (\alpha_i + 1)}{s} \right)^{s} = \left( \frac{n + s}{s} \right)^{s}.$$

Let us consider $f(x) = (\frac{n+x}{x})^x$ function on $\mathbb{R}$.
Taking derivative, we get that
$$\begin{align*} f'(x) &= (1 + \frac{n}{x})^x \ln(1 + \frac{n}{x}) - \frac{n}{x}(1 + \frac{n}{x})^{n-1} = \\ &= (1 + \frac{n}{x})^{x-1} \cdot \left( (1 + \frac{n}{x}) \ln(1 + \frac{n}{x}) - \frac{n}{x} \right). \end{align*}$$

Observe that $x \le \frac{n}{2} \Leftrightarrow \frac{n}{x} \ge 2$. So $\ln(1 + \frac{n}{x}) \ge \ln 3 > 1$, thus we concluded that $(1 + \frac{n}{x}) \ln(1 + \frac{n}{x}) - \frac{n}{x} > 1 > 0$. So $f$ is increasing function on $(0; n/2]$. Finally, the $f(x)$ function get maximum value the point $x = \frac{n}{2}$. Hence we can see that $f(\frac{n}{2}) = 3^{n/2} < 2^n$. Proof is completed.