10. (2003 National Training Team Selection Contest Question) Let $A \subseteq\{0,1,2, \cdots, 29\}$ satisfy: for any integer $k$ and any numbers $a, b$ (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.
10. The desired set $A$ is $\{3l+2 \mid 0 \leqslant l \leqslant 9\}$.
Let $A$ satisfy the conditions of the problem and $|A|$ be maximized. Since the product of two consecutive integers, when divided by 30, leaves a remainder of $0, 2, 6, 12, 20$, or 26, for any $a \in A$, we have $2a \neq 0, 2, 6, 12, 20, 26 \pmod{30}$, which means $a \neq 0, 1, 3, 6, 10, 13, 15, 16, 18, 21, 25, 28$. 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 partitioned into the following 10 subsets, each of which can contain at most one element of $A$: $\{2, 4\}, \{5, 7\}, \{8, 12\}, \{9, 11\}, \{14, 22\}, \{17, 19\}, \{20\}, \{23, 27\}, \{24, 26\}, \{29\}$. Hence, $|A| \leqslant 10$.
If $|A| = 10$, then each subset must contain exactly one element of $A$, so $20 \in A$ and $29 \in A$.
From $20 \in A$, we know $12 \notin A$, thus $8 \in A$, which implies $4 \notin A$, $22 \notin A$, and $24 \notin A$. Therefore, $2 \in A$, $14 \in A$, and $26 \in A$. From $29 \in A$, we know $7 \notin A$ and $27 \notin A$, thus $5 \in A$ and $23 \in A$, which implies $9 \notin A$ and $19 \notin A$. Therefore, $11 \in A$ and $17 \in A$. In summary, $A = \{2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}$, and this $A$ indeed satisfies the requirements.