Example 3 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. Try to find all $A$ with the maximum number of elements. (2003 China National Training Team Selection Exam Problem)
Solve for $A=\{3 t+2 \mid 0 \leqslant t \leqslant 9, t \in \mathbf{Z}\}$.
Let $A$ satisfy the conditions in the problem and $|A|$ be maximized. Since for two consecutive integers $a, a+1$, we have $a(a+1) \equiv 0,2,6,12,20,26(\bmod 30)$. Therefore, for any $a \in A$, taking $b=a, k=0$, we know that $2 a \neq 0,2,6,12,20,26(\bmod 30)$, i.e., $a \neq 0,1,3,6,10,13,15,16,18,21,25,28(\bmod 30)$. Thus, $A \subseteq M=\{2,4,5,7,8,9,11,12,14,17,19,20,22,23,24,26,27,29\}$, and $M$ can be decomposed into the union of the following 10 subsets: $A_{1}=\{2,4\}, A_{2}=\{5,7\}, A_{3}=\{8,12\}, A_{4}=\{11,9\}, A_{5}=\{14,22\}$, $A_{6}=\{17,19\}, A_{7}=\{20\}, A_{8}=\{23,27\}, A_{9}=\{26,24\}, A_{10}=\{29\}$. Each subset $A_{i}$ can contain at most one element from $A$, hence $|A| \leqslant 10$.

If $|A|=10$, then each subset $A_{i}$ contains exactly one element from $A$, thus $20 \in A$, $29 \in A$. From $20 \in A$, we know $12 \notin A, 22 \notin A$, so $8 \in A, 14 \in A$, which means $4 \notin A$, $24 \notin A$. Therefore, $2 \in A, 26 \in A$. From $29 \in A$, we know $7 \notin A, 27 \notin A$, so $5 \in A$, $23 \in A$, which means $9 \notin A, 19 \notin A$, thus $11 \in A, 17 \in A$.

In summary, the set $A=\{2,5,8,11,14,17,20,23,26,29\}$, and it is verified that $A$ meets the requirements.