Example 1: For $n$ consecutive positive integers, if each number is written in its standard prime factorization form, and each prime factor is raised to an odd power, such a sequence of $n$ consecutive positive integers is called a "consecutive $n$ odd group" (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \times 3^{1}$, then $22, 23, 24$ form a consecutive 3 odd group). The maximum possible value of $n$ in a consecutive $n$ odd group is $\qquad$ [1]
【Analysis】Notice that, in a connected $n$-singular group, if there exists a multiple of 4, then by the definition of a connected $n$-singular group, it must be a multiple of 8.

Let this number be $2^{k} A\left(k, A \in \mathbf{N}_{+}, k \geqslant 3, A\right.$ is an odd number). Then $2^{k} A+4$ and $2^{k} A-4$ are both numbers that are congruent to 4 modulo 8.
Therefore, the power of 2 in their prime factorizations is even.
Thus, a connected $n$-singular group cannot contain two multiples of 4 simultaneously. However, in any 8 consecutive positive integers, there must be a number that is congruent to 0 modulo 8 and a number that is congruent to 4 modulo 8. Therefore, in any 8 consecutive positive integers, there must be two multiples of 4.
Hence, $n \leqslant 7$.
Furthermore, $29,30, \cdots, 35$ is a connected 7-singular group, so $n_{\max }=7$.