15 Given the set $P$ consists of positive integers not exceeding 2012, i.e., $P=$ $\{1,2,3, \cdots, 2012\}$. The set $A$ is a subset of set $P$, and the symbol $|A|$ represents the number of elements in set $A$, while $S(A)$ represents the sum of all elements in set $A$.
(1) If the difference between any two numbers in set $A$ is not a multiple of 101, find the maximum value of $|A|$;
(2) If the difference between any two numbers in set $A$ is not a multiple of 101, and the sum of any two numbers is also not a multiple of 101, find the maximum value of $|A|$;
(3) If the difference between any two numbers in set $A$ is not a multiple of 101, the sum of any two numbers is also not a multiple of 101, and $S(A)=2012$, find the maximum value of $|A|$.
(15) (1) Construct the following 101 sets: $A_{i}=\{x \mid x \equiv i(\bmod 101)$, and $x \in P\}(i=0,1,2, \cdots, 100)$.

It is easy to see that $A_{i}$ are all subsets of $P$, and the difference between any two numbers in $A_{i}$ is a multiple of 101.

If $|A| \geqslant 102$, then by the pigeonhole principle, set $A$ contains two numbers from one of the above 101 sets, and the difference between these two numbers is a multiple of 101, which does not meet the requirement. Therefore, $|A| \leqslant 101$.

Furthermore, a set formed by taking one element from each of the above 101 sets, such as the set $A=\{1,2,3, \cdots, 101\}$, clearly meets the requirement, and $|A|=101$.
Thus, the maximum value of $|A|$ is 101.
(2) Construct the following 51 sets:
$A_{i}=\{x \mid x \equiv i(\bmod 101)$, or $x \equiv 101-i(\bmod 101)$, and $x \in P\}$ $(i=0,1,2, \cdots, 50)$.

It is easy to see that $A_{i}$ are all subsets of $P$, and the difference or sum of any two numbers in $A_{i}$ is a multiple of 101.

If $|A| \geqslant 52$, then by the pigeonhole principle, set $A$ contains two numbers from one of the above 51 sets, and the difference or sum of these two numbers is a multiple of 101, which does not meet the requirement. Therefore, $|A| \leqslant 51$.

Furthermore, a set formed by taking one element from each of the above 51 sets, such as the set $A=\{1,2,3, \cdots, 50,101\}$, clearly meets the requirement, and $|A|=51$.
Thus, the maximum value of $|A|$ is 51.
(3) Take the set $A=\{1,2,3, \cdots, 35,65,37,38, \cdots, 49,51,7 \times 101\}$, i.e., in the set $A$ constructed in (2), replace 36 with 65, 50 with 51, and 101 with $7 \times 101$.

Since $65 \equiv 101-36(\bmod 101), 51 \equiv 101-50(\bmod 101)$, the difference between any two numbers in set $A$ is not a multiple of 101, and the sum of any two numbers is also not a multiple of 101.
Furthermore,
$$
\begin{aligned}
S(A) & =1+2+3+\cdots+50+(65-36)+1+707 \\
& =1275+29+1+707=2012,
\end{aligned}
$$

Thus, $A$ meets the requirement. Therefore, the maximum value of $|A|$ is 51.