Example 2 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 $a, b, c$ in $S$ such that $a+b, b+c, c+a$ are all in the subset $A$.
(2012, China Mathematical Olympiad)
【Analysis】The condition of the problem is equivalent to being able to find three numbers $x, y, z$ in the subset $A$ such that
$$
\frac{-x+y+z}{2}, \frac{x-y+z}{2}, \frac{x+y-z}{2}
$$

are three distinct positive integers. For this, $x, y, z$ must be distinct, satisfy the triangle inequality, and their sum must be even.

A simple estimation shows that taking all the odd numbers can yield a 1006-element subset, where the sum of any three elements is odd, which is a larger counterexample to the required subset $A$. Further, it is found that adding the element 2 can result in a 1007-element counterexample. Therefore, we conjecture that $k=1008$.
Next, we prove this conjecture.
Here, we use a simple principle in combinatorial extremal problems: "The smallest $k$ for which all subsets $A$ of at least $k$ elements satisfy condition $P$" minus 1 is exactly "the largest $n$ for which there exists an $n$-element subset that does not satisfy condition $P$." Let condition $P$ be: the sum of three different numbers is even and satisfies the triangle inequality.

First, we use the pigeonhole principle to prove: any 1008-element subset $A$ of $S$ always contains three elements that satisfy condition $P$.

Since $|A|>\frac{|S|}{2}$ ensures that some pigeonhole $T$ contains more than $\frac{|T|}{2}$ elements of $A$, thus, the pigeonhole $T$ should satisfy the condition: any more than $\frac{|T|}{2}$ elements in $T$ contain three elements that satisfy condition $P$. Then $\frac{|T|}{2} \geqslant 3$, i.e., $|T| \geqslant 6$.

Next, we prove: from the set $\{3,4, \cdots, 8\}$, any four numbers contain three numbers that satisfy condition $P$.
The simplest and most direct proof is by enumeration.
List all three-element subsets of $\{3,4, \cdots, 8\}$ that satisfy condition $P$, to prove that all four-element subsets contain some three-element subset that satisfies condition $P$, it is only necessary to prove that the complements of all three-element subsets that satisfy condition $P$ contain all possible two-element subsets. Thus, find the complements of all three-element subsets that satisfy condition $P$, and list all two-element subsets of the complements, and then, check that all two-element subsets of the six-element set. Using vertices to represent the six elements and edges to represent the two-element subsets, it is easy to verify this by drawing a graph.
By translation, it can be proven that for any positive integer $n$,
$$
\{2 n+3,2 n+4, \cdots, 2 n+8\}
$$

is also a pigeonhole that satisfies the condition: the translation step is even, ensuring that the parity of the numbers at corresponding positions is the same as in the original pigeonhole, and the condition that the sum is even remains unchanged; all numbers increase, and the corresponding triangle inequality condition weakens.

Divide $S=\{1,2, \cdots, 2012\}$ into $\{1,2\}$, $\{6 n+3,6 n+4, \cdots, 6 n+8\}(n=0,1, \cdots, 334)$.

When $A \subset S$ and $A$ contains at least 1008 elements, $A$ contains at least four numbers in some pigeonhole $\{6 n+3,6 n+4, \cdots, 6 n+8\}$, thus, there are three numbers that satisfy condition $P$, and the required $a, b, c$ can be solved.
Therefore, the smallest $k=1008$.