Proof: Let the $n$ points inserted in the middle of line segment $AB$ be $A_{1}, A_{2}, \cdots A_{n}$. Suppose there are $k$ standard segments, and the points $A_{i}$ are associated with numbers $a_{i}$ as follows:
$$\quad a_{i}=\left\{\begin{array}{l} 1, \text { if point } A_{i} \text { is red } \\ -1, \text { if point } A_{i} \text { is blue } \end{array}\right.$$

Then, by the problem statement, $a_{A} \cdot a_{B}=\cdots-1$, and
$$\begin{array}{l} a_{A} \cdot a_{B}==a_{A} \cdot a_{B} \cdot a_{1}^{2} \cdot a_{2}^{2} \cdots a_{n}^{2} \\ =\left(a_{A} a_{1}\right)\left(a_{1} a_{2}\right)\left(a_{2} a_{3}\right) \cdots\left(a_{n} a_{B}\right) \\ =(-1)^{k}, \end{array}$$

Therefore, $(-1)^{A}=-1$, which means $k$ is odd.