2 Let $A \cong\{0,1,2, \cdots, 29\}$ satisfy: for any integer $k$ and any numbers $a, b$ in $A$ (where $a, b$ can be the same), $a+b+30 k$ is not the product of two consecutive integers. Determine all $A$ with the maximum number of elements. (Chen Yonggao)
The required $A$ is $\{3 l+2 \mid 0 \leqslant l \leqslant 9\}$.
Let $A$ satisfy the conditions in the problem and $|A|$ be the maximum. 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 $2 a \neq 0,2,6,12,20,26(\bmod 30)$, i.e., $a \neq 0,1,3,6,10,13,15,16,18,21,25,28$, thus, $A \subseteq\{2,4,5,7,8,9,11,12,14,17,19,20,22,23,24,26,27,29\}$, the latter set can be decomposed into the union of the following 10 subsets, each of which can contain at most one element in $A$: $\{2,4\},\{5,7\},\{8,12\},\{9,11\},\{14,22\},\{17,19\},\{20\},\{23,27\},\{24,26\},\{29\}$, hence $|A| \leqslant 10$.
If $|A|=10$, then each subset contains exactly one element in $A$, therefore, $20 \in A, 29 \in A$.
From $20 \in A$ we know $12 \notin A$, thus $8 \in A$, this way $4 \notin A, 22 \notin A$, $24 \notin A$. Therefore $2 \in A, 14 \in A, 25 \in A$.
From $29 \in A$ we know $7 \notin A, 27 \notin A$, thus $5 \in A, 23 \in A$, this way $9 \notin A, 19 \notin A$, therefore $11 \in A, 17 \in A$.
In summary, $A=\{2,5,8,11,14,17,20,23,26,29\}$, this $A$ indeed meets the requirements.