, 2006

i) Let $A$ be a stubborn subset of $S = \{1, 2, \dots, 2006\}$ consisting of $x$ elements $a_1 < a_2 < \dots < a_x$. Consider the set $B := \{a_2 - a_1, a_3 - a_1, \dots, a_x - a_1\}$. It is a subset of $S$ and consists of $x-1$ elements. As $A$ is a stubborn subset of $S$, $A \cap B = \emptyset$. It implies that $x + (x-1) \leq 2006$, so $x \leq 1003$.

ii) Let $S = \{a_1, a_2, \dots, a_{2006}\}$. Consider the product $P$ of all odd divisors of $\prod_{i=1}^{2006} a_i$. It is easy to see that there exists a prime number $p$ of the form $p = 3r+2$ such that $p$ is a divisor of $3P+2$; $p$ is coprime with every $a_i$ ($i = 1, 2, \dots, 2006$). For each $a \in S$, the sequence $a, 2a, \dots, (p-1)a$ (mod $p$) is a permutation of $1, 2, \dots, p-1$, therefore there exists a set $S_a$ consisting of $r+1$ integers $x$ in $\overline{1, p-1}$ so that $xa\ (\text{mod}\ p)$ belongs to $A = \{r+1, \dots, 2r+1\}$. For each $x \in \overline{1, p-1}$, let $S_x = \{a \in S \mid xa \in A\}$. We have:
$$|S_1| + |S_2| + \dots + |S_{p-1}| = \sum_{a \in S} |A_a| = 2006 \times (r+1)$$
So there exists $x_0$ such that $|S_{x_0}| \geq \frac{2006 \times (r+1)}{3r+1} > 668$. Let $B$ be a subset consisting of 669 elements of $S_{x_0}$ then $B$ is a stubborn subset of $S$. Indeed, if $u, v, w \in B$ ($u$ can be equal to $v$) then $x_0u, x_0v, x_0w \in A$. It is easy to verify that $x_0u + x_0v \neq x_0w\ (\text{mod}\ p)$ therefore $u + v \neq w$.