6. Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a$, $b$, and $c$ in $S$ such that $a+b$, $b+c$, and $c+a$ are all in the set $A$.
6. Let $a<b<c$. Let
$x=a+b, y=a+c, z=b+c$.
Then $x\langle y\langle z, x+y\rangle z$, and $x+y+z$ is even. (1)
Conversely, if there exist $x, y, z \in A$ satisfying property (1), then take
$$
a=\frac{x+y-z}{2}, b=\frac{x+z-y}{2}, c=\frac{y+z-x}{2},
$$

we have $a, b, c \in \mathbf{Z}, 1 \leqslant a<b<c \leqslant 2012$, and
$$
x=a+b, y=a+c, z=b+c.
$$

Thus, the condition of the problem is equivalent to the fact that for any $k$-element subset $A$, there exist $x, y, z \in A$ satisfying property (1).

If $A=\{1,2,3,5,7, \cdots, 2011\}$, then $|A|=$ 1007, and the set $A$ does not contain three elements satisfying property (1). Therefore, $k \geqslant 1008$.

Next, we prove that any 1008-element subset contains three elements satisfying property (1).
We will prove a more general conclusion:
For any integer $n(n \geqslant 4)$, any $(n+2)$-element subset of the set $\{1,2, \cdots, 2 n\}$ contains three elements satisfying property (1).
We use induction on $n$.
When $n=4$, let $A$ be a six-element subset of $\{1,2, \cdots, 8\}$. Then $A \cap\{3,4, \cdots, 8\}$ has at least 4 elements.

If $A \cap\{3,4, \cdots, 8\}$ contains three even numbers, then $4, 6, 8 \in A$ and they satisfy property (1);

If $A \cap\{3,4, \cdots, 8\}$ contains exactly two even numbers, then it also contains at least two odd numbers. Taking these two odd numbers, at least two even numbers from $4, 6, 8$ can form a three-element set satisfying property (1), since there are at least two even numbers, there exist three numbers satisfying property (1);

If $A \cap\{3,4, \cdots, 8\}$ contains exactly one even number, then it contains all three odd numbers, and this even number with 5, 7 forms a three-element set satisfying property (1).
Therefore, the conclusion holds when $n=4$.
Assume the conclusion holds for $n(n \geqslant 4)$, consider the case for $n+1$.

Let $A$ be an $(n+3)$-element subset of $\{1,2, \cdots, 2 n+2\}$. If $|A \cap\{1,2, \cdots, 2 n\}| \geqslant n+2$, then by the induction hypothesis, the conclusion holds. Thus, we only need to consider the case
$$
|A \cap\{1,2, \cdots, 2 n\}| = n+1,
$$

and $2 n+1, 2 n+2 \in A$.
In this case, if $\{1,2, \cdots, 2 n\}$ contains an odd number $x$ greater than 1 in the set $A$, then $x, 2 n+1, 2 n+2$ form a three-element set satisfying property (1);

If $\{1,2, \cdots, 2 n\}$ contains no odd numbers greater than 1 in the set $A$, then
$$
A \subset\{1,2,4,6, \cdots, 2 n, 2 n+1,2 n+2\},
$$

and the latter has exactly $n+3$ elements, hence
$$
A=\{1,2,4,6, \cdots, 2 n, 2 n+1,2 n+2\},
$$

in this case, $4, 6, 8 \in A$ satisfy property (1).
In conclusion, the smallest $k$ is 1008.