It can be the same), $a+b+30k$ are not the product of two consecutive integers. Try to determine the set $A$ with the smallest number of elements.



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It can be the same), $a+b+30k$ are not the product of two consecutive integers. Try to determine the set $A$ with the smallest number of elements.
First, $0+0=0 \times 1, 1+1=1 \times 2, 3+3=2 \times 3, 6+6=3 \times 4, 10+10=4 \times 5, 15+15=5 \times 6, 21+21=6 \times 7, 28+28=7 \times 8$. Therefore, $0, 1, 3, 6, 10, 15, 21, 28 \notin A$. Next,
$$
\begin{array}{l}
13+13+30-7 \times 8, 16+16-30=1 \times 2, \\
18+18-30=2 \times 3, 25+25-30=4 \times 5 .
\end{array}
$$

Next, $13, 16, 18, 25 \notin A$. Finally,
$$
\begin{array}{l}
2+4=2 \times 3, 5+7=3 \times 4, 8+12=4 \times 5, \\
9+11=4 \times 5, 14+22-30=2 \times 3, \\
17+19-30=2 \times 3, 23+27-30=4 \times 5, \\
24+26-30=4 \times 5 .
\end{array}
$$

Therefore, for the 8 pairs of numbers 2 and 4, 5 and 7, 8 and 12, 9 and 11, 14 and 22, 17 and 19, 23 and 27, 24 and 26, at most one number from each pair can be in $A$. Thus, $A$ can contain at most 8 of these numbers, plus 20 and 29, so $|A| \leqslant 10$.
On the other hand, $n(n+1) \equiv 0, 2 (\bmod 3)$. Let
$$
A=\{2, 5, 8, 11, 14, 17, 20, 23, 26, 29\},
$$

then for any two numbers $a, b$ in $A$ and any integer $k, a+b+30k \equiv 1 (\bmod 3)$, so $a+b+30k$ is not the product of two consecutive integers.
Therefore, (1) is the $A$ with the maximum number of elements, which is 10.