10 . Rearrange the sequence of consecutive natural numbers $1,2, \cdots, n(n \geqslant 3)$ into a new sequence $a_{1}, a_{2}, \cdots, a_{n}$. If $\left(a_{1}-1\right)\left(a_{2}-2\right) \cdots\left(a_{n}-n\right)$ is exactly an odd number, then $n$ ( ).
(A) must be even
(B) must be odd
(C) can be either odd or even
(C) must be $2^{m}-1(m$ is odd $)$
10. A.

From $\left(a_{1}-1\right)\left(a_{2}-2\right) \cdots\left(a_{n}-n\right)$ being odd, we know that $a_{1}-1, a_{2}-2, \cdots, a_{n}-n$ are all odd. If $n$ is odd, then
$$
\left(a_{1}-1\right)+\left(a_{2}-2\right)+\cdots+\left(a_{n}-n\right)
$$

is odd. However,
$$
\begin{array}{l}
\left(a_{1}-1\right)+\left(a_{2}-2\right)+\cdots+\left(a_{n}-n\right) \\
=\left(a_{1}+a_{2}+\cdots+a_{n}\right)-(1+2+\cdots+n)=0,
\end{array}
$$

which is a contradiction. Therefore, $n$ is even.