Let $n \geq 3$ be an odd number and suppose that each square in an $n \times n$ chessboard is colored either black or white. Two squares are considered adjacent if they are of the same color and share a common vertex. Two squares $a$ and $b$ are considered connected if there exists a sequence of squares $c_1, \ldots, c_k$ with $c_1 = a$ and $c_k = b$ such that $c_i$ and $c_{i+1}$ are adjacent for $i = 1, 2, \ldots, k-1$.

We aim to find the maximal number $M$ such that there exists a coloring admitting $M$ pairwise disconnected squares.

To solve this problem, we need to consider the structure of the chessboard and the properties of the coloring. The key insight is to analyze the number of disjoint maximal monochromatic components in the board.

For a general $(2m+1) \times (2n+1)$ board, we can prove that the maximal number of disjoint components is given by:
$$M = (m+1)(n+1) + 1.$$

This result can be established through induction and careful analysis of the board's configuration. The proof involves considering different types of configurations and using combinatorial arguments to bound the number of components.

Hence, the maximal number $M$ of pairwise disconnected squares in an $n \times n$ chessboard, where $n$ is an odd number, is:
$$M = \left(\frac{n+1}{2}\right)^2 + 1.$$

The answer is: \boxed{\left(\frac{n+1}{2}\right)^2 + 1}.