Example 8 (2003 China National Team Training Selection for IMO) Let $A \subseteq \{0,1,2, \cdots, 29\}$, satisfying: for any integer $k$ and any numbers $a, b \in A$ (where $a, b$ can be the same), $a+b+30k$ is not the product of two consecutive integers. Determine all $A$ with the maximum number of elements.
Let $A$ satisfy the conditions of the problem and $|A|$ be the largest.
Since the product of two consecutive integers, when divided by 30, leaves a remainder of $0,2,6,12,20,26$, then for any $a \in A$, we have
$$
2 a \neq 0,2,6,12,20,26(\bmod 30),
$$

which means $a \neq 0,1,3,6,10,13,15,16,18,21,25,28(\bmod 30)$.
Therefore, $A \subseteq\{2,4,5,7,8,9,11,12,14,17,19,20,22,23,24,26,27,29\}$.
The latter set can be decomposed into the union of the following 10 subsets, each of which contains at most one element of $A$: $\{2,4\},\{5,7\},\{8,12\},\{11,9\},\{14,22\},\{17,19\},\{20\},\{23,27\},\{26,14\},\{29\}$.
Thus, $|A| \leqslant 10$.
If $|A|=10$, then each subset contains exactly one element of $A$, so $20 \in A, 29 \in A$.
From $20 \in A$, we know $12 \notin A, 22 \notin A$, thus $8 \in A, 14 \in A$. This implies $4 \notin A, 24 \notin A$, so $2 \in A, 26 \in A$. From $29 \in A$, we know $7 \notin A, 27 \notin A$, thus $5 \in A, 23 \in A$. This implies $9 \notin A, 19 \notin A$, so $11 \in A, 17 \in A$. In summary, $A=\{2,5,8,11,14,17,20,23,26,29\}=\{3 l+2 \mid 0 \leqslant l \leqslant 9\}$, and this $A$ indeed satisfies the requirements.