2. Let $A_{1} A_{2} \cdots A_{101}$ be a regular 101-gon. Each vertex is colored either red or blue. Let $N$ be the number of obtuse triangles that satisfy the following conditions: the three vertices of the triangle are vertices of the 101-gon, the two acute vertices have the same color, and the color of the obtuse vertex is different from that of the acute vertices. Find:
(1) the maximum possible value of $N$;
(2) the number of different coloring methods that make $N$ achieve its maximum value (for two coloring methods, if the color on some $A_{i}$ is different, they are considered different coloring methods). (Tian Zhenhua, problem contributor)
2. Let $x_{i}=0$ or 1 represent $A_{i}$ being red or blue, respectively. For an obtuse triangle $\triangle A_{i-a} A_{i} A_{i+b}$, where vertex $A_{i}$ is the obtuse angle vertex, i.e., $a+b \leqslant 50$.
The coloring of the three vertices satisfies the condition if and only if
$$
\left(x_{i}-x_{i-a}\right)\left(x_{i}-x_{i+b}\right)=1 \text {, }
$$
otherwise, it equals 0, where the indices are taken modulo 101.
$$
\text { Hence } N=\sum_{i=1}^{101} \sum_{a, b}\left(x_{i}-x_{i-a}\right)\left(x_{i}-x_{i+b}\right) \text {, }
$$
where, “ $\sum_{(a, b)}$ ” is the sum over all ordered pairs of positive integers $(a, b)$ such that $a+b \leqslant 50$, and there are
$$
49+48+\cdots+1=1225 \text { (pairs). }
$$
Substituting equation (1) into equation (2) gives
$$
\begin{array}{l}
N=\sum_{i=1}^{101} \sum_{a, b}\left(x_{i}-x_{i-a}\right)\left(x_{i}-x_{i+b}\right) \\
=\sum_{i=1}^{101} \sum_{a, b}\left(x_{i}^{2}-x_{i} x_{i-a}-x_{i} x_{i+b}+x_{i-a} x_{i+b}\right) \\
=1225 \sum_{i=1}^{101} x_{i}^{2}+\sum_{i=1}^{101} \sum_{k=1}^{\infty}[k-1-2(50-k)] x_{i} x_{i+k} \\
=1225 n+\sum_{i=1}^{101} \sum_{k=1}^{\infty}(3 k-101) x_{i} x_{i+k},
\end{array}
$$
where $n$ is the number of blue vertices.
For any two vertices $A_{i} A_{j}(1 \leqslant i \leqslant j \leqslant 101)$, let
$$
d\left(A_{i}, A_{j}\right)=d\left(A_{j}, A_{i}\right)=\min \{j-i, 101-j+i\} .
$$
Let $B \subseteq\left\{A_{1}, A_{2}, \cdots, A_{101}\right\}$ be the set of all blue vertices. Then equation (3) can be written as
$$
N=1225 n-101 \mathrm{C}_{n}^{2}+3 \sum_{|P, Q| \leq B} d(P, Q),
$$
where $\{P, Q\}$ runs over all two-element subsets of $B$.
Assume $n$ is even; otherwise, change the color of each vertex, and the value of $N$ remains unchanged.
Let $n=2 t(0 \leqslant t \leqslant 50)$, and reindex all blue vertices in a clockwise direction starting from some point as $P_{1}, P_{2}, \cdots, P_{2 t}$. Hence,
$$
\begin{array}{l}
\sum_{|P, Q| \leq B} d(P, Q) \\
=\sum_{i=1}^{t} d\left(P_{i}, P_{i+t}\right)+\frac{1}{2} \sum_{i=1}^{t} \sum_{j=1}^{t-1}\left(d\left(P_{i}, P_{i+j}\right)+\right. \\
\left.d\left(P_{i+j}, P_{i+t}\right)+d\left(P_{i+t}, P_{i-j}\right)+d\left(P_{i-j}, P_{i}\right)\right) \\
\leqslant 50 t+\frac{101}{2} t(t-1),
\end{array}
$$
where the indices of $P_{i}$ are taken modulo $2 t$, and the inequality $d\left(P_{i}, P_{i+1}\right) \leqslant 50$ and
$$
\begin{array}{l}
d\left(P_{i}, P_{i+j}\right)+d\left(P_{i+j}, P_{i+t}\right)+ \\
d\left(P_{i+t}, P_{i-j}\right)+d\left(P_{i-j}, P_{i}\right) \leqslant 101
\end{array}
$$
are used.
Combining equations (4) and (5) gives
$$
\begin{array}{l}
N \leqslant 1225 n-101 \mathrm{C}_{n}^{2}+3\left[50 t+\frac{101}{2} t(t-1)\right] \\
=-\frac{101}{2} t^{2}+\frac{5099}{2} t .
\end{array}
$$
The right-hand side of the above equation achieves its maximum value when $t=25$.
Thus, $N \leqslant 32175$.
We need to consider the necessary and sufficient conditions for $N=32175$.
First, $t=25$, i.e., there are exactly 50 blue points.
Second, for $1 \leqslant i \leqslant t$, $d\left(P_{i}, P_{i+t}\right)=50$.
When $d\left(P_{i}, P_{i+1}\right)=50$, the equality in equation (6) must hold.
Therefore, the number of ways to choose 50 blue points such that $N$ achieves its maximum value is the number of ways to choose 25 diagonals from the longest 101 diagonals such that any two diagonals intersect internally (equivalent to having no common vertices).
Connect $A_{i}(i=1,2, \cdots, 101)$ with $A_{i+50}$ to form a graph $G$.
Note that 50 and 101 are coprime.
Thus, $1+50 n(0 \leqslant n \leqslant 100)$ forms a complete residue system modulo 101, i.e., graph $G$ is a cycle with exactly 101 edges.

Therefore, the number of ways to color 50 vertices blue such that $N$ achieves its maximum value, $S$, is the number of ways to select 25 edges from graph $G$ such that no two edges are adjacent in $G$.

Now, fix an edge $e$ in graph $G$. The number of ways to choose 25 edges including edge $e$ is $\mathrm{C}_{75}^{24}$, and the number of ways to choose 25 edges excluding edge $e$ is $\mathrm{C}_{76}^{25}$. Hence, the number of ways is
$$
S=\mathrm{C}_{75}^{24}+\mathrm{C}_{76}^{25} \text {. }
$$
Considering the case of 50 red points also has $S$ ways of coloring. Therefore, the number of coloring methods that make $N$ achieve its maximum value is $2 S$.
In summary, the maximum value of $N$ is 32175, and the corresponding number of coloring methods is $2 S=2\left(\mathrm{C}_{75}^{24}+\mathrm{C}_{76}^{25}\right)$.