Given a set of 100 points in the plane, we want to determine the maximum number of harmonic pairs, where a pair $(A, B)$ of points is considered harmonic if $1 < d(A, B) \leq 2$ and $d(A, B) = |x_1 - x_2| + |y_1 - y_2|$.

To solve this problem, we can transform the distance function to make it easier to handle. By rotating the plane by 45 degrees, we change the coordinates of a point $P = (x, y)$ to $P' = (x - y, x + y)$. Under this transformation, the Manhattan distance $d(P, Q)$ becomes $d'(P', Q') = \max \{ |P'_x - Q'_x|, |P'_y - Q'_y| \}$.

We claim that the maximum number of harmonic pairs is $\frac{3 \times 100^2}{4 \times 2} = 3750$. To achieve this bound, we can place 25 points each in small neighborhoods around the four points $(\pm \frac{1.0201082102011209}{2}, \pm \frac{1.0201082102011209}{2})$.

To prove that this is the maximum number, we construct a graph $G$ with 100 vertices, where two vertices are connected if the corresponding points are harmonic. We need to show that $G$ has no $K_5$ (complete graph on 5 vertices).

Claim: $G$ has no $K_5$.

Proof: Consider the following two facts:
1. If a coloring of the edges of $K_5$ with two colors does not produce a monochromatic triangle, then it must have a monochromatic cycle of length 5.
2. It is impossible to find three real numbers $A, B, C$ such that all points $(A, 0), (B, 0), (C, 0)$ are mutually harmonic.

For each edge $PQ$ in $G$, color the edge red if $\max \{ |P_x - Q_x|, |P_y - Q_y| \} = |P_x - Q_x|$, or blue otherwise. Suppose, for contradiction, that there is a $K_5$ in $G$ with points $A, B, C, D, E$. By fact 2, it has no monochromatic triangle, so by fact 1, it has a monochromatic cycle of length 5. Without loss of generality, assume the cycle is red, and let it be $A \rightarrow B \rightarrow \cdots \rightarrow E$.

If $\max(A_y, B_y, C_y, D_y, E_y) - \min(A_y, B_y, C_y, D_y, E_y) > 2$, we have a contradiction because the extreme points would not be harmonic. Therefore, $\max(A_y, B_y, C_y, D_y, E_y) - \min(A_y, B_y, C_y, D_y, E_y) \leq 2$. Assume $\min(A_y, B_y, C_y, D_y, E_y) = A_y = 0$, so $\max(A_y, B_y, C_y, D_y, E_y) \leq 2$. Thus, $A_y, B_y, C_y, D_y, E_y \in [0, 2]$.

Color the vertices with ordinate in $[0, 1]$ black and those in $(1, 2]$ white. Traversing $A \rightarrow B \rightarrow \cdots \rightarrow E$ changes the color of the interval each time, implying the odd cycle is bipartite, which is a contradiction.

By Turan's theorem, the strictest bound possible for the number of edges in $G$ without a $K_5$ is $\frac{3 \times 100^2}{4 \times 2} = 3750$.

The answer is $\boxed{3750}$.