Define $x_i = 0$ or $1$ depending on whether $A_i$ is red or blue. For obtuse triangle $A_{i-a}A_iA_{i+b}$ (vertex $A_i$ is the vertex of the obtuse angle, i.e. $a + b \le 50$), these three vertices satisfy the conditions of the problem if and only if
$$(x_i - x_{i-a})(x_i - x_{i+b}) = 1, \qquad \textcircled{1}$$
otherwise equals $0$, here the subscript is modulo $101$. Thus
$$N = \sum_{i=1}^{101} \sum_{(a,b)} (x_i - x_{i-a})(x_i - x_{i+b}),$$
here $\sum_{(a,b)}$ stands for the summation of all positive integer pairs $(a, b)$ satisfying $a + b \le 50$, there are $49 + 48 + \cdots + 1 = 1225$ such positive integer pairs. Expand $\textcircled{1}$, we have
$$\begin{align*} N &= \sum_{i=1}^{101} \sum_{(a,b)} (x_i - x_{i-a})(x_i - x_{i+b}) \\ &= \sum_{i=1}^{101} \sum_{(a,b)} (x_i^2 - x_i x_{i-a} - x_i x_{i+b} + x_{i-a} x_{i+b}) \\ &= 1225 \sum_{i=1}^{101} x_i^2 + \sum_{i=1}^{101} \sum_{k=1}^{50} (k-1-2(50-k)) x_i x_{i+k} \end{align*}$$

$$= 1225n + \sum_{i=1}^{101} \sum_{k=1}^{50} (3k-101)x_i x_{i+k}, \qquad \textcircled{2}$$
here $n$ is the number of the blue vertices. For any two vertices $A_i, A_j$,
$1 \le i \le j \le 101$, let
$$d(A_i, A_j) = d(A_j, A_i) = \min\{j - i, 101 - j + i\}.$$
Let $B \subseteq \{A_1, A_2, \dots, A_{101}\}$ be the set of all blue vertices, then $\textcircled{2}$ can be written as
$$N = 1225n - 101C_n^2 + 3 \sum_{\{P,Q\} \subseteq B} d(P, Q), \qquad \textcircled{3}$$
where $\{P, Q\}$ runs through all the two-element subsets of $B$. Without loss of generality, we assume $n$ is even. Otherwise, color all the vertices the opposite, the value of $N$ does not change. Write $n = 2t$, $0 \le t \le 50$, from one point, renumber all the blue vertices as $P_1, P_2, \dots, P_{2t}$ clockwise. Then
$$\sum_{\{P,Q\} \subseteq B} d(P, Q) = \sum_{i=1}^t d(P_i, P_{i+t}) + \frac{1}{2} \sum_{i=1}^t \sum_{j=1}^{t-1} (d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i)) \le 50t + \frac{101}{2} t(t-1), \qquad \textcircled{4}$$
here the subscript of $P_i$ is modulo $2t$, and we use the inequality $d(P_i, P_{i+t}) \le 50$ and
$$d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) \le 101. \qquad \textcircled{5}$$
Combine $\textcircled{3}$ and $\textcircled{4}$, we have
$$N \le 1225n - 101C_n^2 + 3 \left( 50t + \frac{101}{2} t(t-1) \right) = -\frac{101}{2} t^2 + \frac{5099}{2} t. \qquad \textcircled{6}$$
The right hand side of $\textcircled{6}$ attains its maximum value when $t = 25$, thus $N \le 32175$.

The number of ways to choose $50$ blue vertices such that $N$ achieves maximum is equal to the number of ways to choose $25$ diagonals from the longest $101$ diagonals such that any two diagonals do not have common vertices.

Edging $A_i$ and $A_{i+50}$, $i = 1, 2, \dots, 101$, we get a graph $G$. Notice that $50$ and $101$ are relatively prime, so $1 + 50n$ ($0 \le n \le 100$) form a complete residue system modulo $101$, i.e. $G$ is a circle with $101$ edges. Therefore, the number of ways (written as $S$) to color $50$ vertices blue and $N$ achieves maximum, is equal to the number of ways to choose $25$ edges of $G$, such that any two of them do not have common vertices. Now, fix one edge $e$ of $G$. Since the number of ways to choose edges having $e$ is $C_{75}^{24}$, not having $e$ is $C_{76}^{25}$, so $S = C_{75}^{24} + C_{76}^{25}$. Similarly, the number of ways to have $50$ red vertices is also $S$, thus the number of ways to color the vertices such that the largest value of $N$ is achieved is $2S$.

In a word, the largest possible value of $N$ is $32175$, the number of ways to color is $2S = 2(C_{75}^{24} + C_{76}^{25})$.