1. If the irreducible proper fraction $\frac{a}{b}$ satisfies $a+b=2007$ $\left(a, b \in \mathbf{N}_{+}\right)$, then it is called an "ideal fraction". Then, the number of ideal fractions is ( ).
(A) 337
(B) 666
(C) 674
(D) 1332
- 1.B.
Given $a+b=2007$ and $a1$, then $d \mid 2007$.
Since $2007=3^{2} \times 223$, we have $3 \mid d$ or $223 \mid d$.
In the set $M=\{1,2, \cdots, 1003\}$, the numbers divisible by 3 are $\left[\frac{1003}{3}\right]=334$, the numbers divisible by 223 are $\left[\frac{1003}{223}\right]=4$, and the numbers divisible by $3 \times 223$ are $\left[\frac{1003}{669}\right]=1$. Therefore, the numbers in $M$ that are not coprime with 2007 are $334+4-1=337$. Thus, the numbers in $M$ that are coprime with 2007 are $1003-337=666$. Hence, there are 666 ideal fractions.