Example 2 Find the number of integers in the set $\left\{\left.\frac{2015[a, b]}{a+b} \right\rvert\, a 、 b \in \mathbf{Z}_{+}\right\}$.
Let $d=(a, b)$, and $a=A d, b=B d$, where $A$ and $B$ are coprime positive integers.
Since $[a, b]=A B(a, b)$, then
$(a+b)|2015[a, b] \Leftrightarrow(A+B)| 2015 A B$.
Because $(A, B)=1$, we have
$(A+B, A)=1,(A+B, B)=1$,
$(A+B, A B)=1$.
Thus, $(A+B) \mid 2015$.
For a fixed divisor $k$ of 2015 greater than 1, and $k$ is odd, let the number of positive integers less than $k$ and coprime to $k$ be $\varphi(k)$. Suppose these numbers are
$$
1=c_{1}<c_{2}<\cdots<c_{\varphi(k)} \leq k-1
$$
and
$$
c_{1}+c_{2}+\cdots+c_{\varphi(k)}=\frac{\varphi(k)(k-1)}{2} \quad \text{and} \quad \varphi(1)} \varphi(k)=n-1$.
To prove that in the set $\left\{\frac{1}{n}, \frac{2}{n}, \cdots, \frac{n-1}{n}\right\}$, when the fractions are simplified to their lowest terms, the denominators can only be the divisors $k$ of $n$ that are greater than or equal to 2, and the number of fractions with denominator $k$ is $\varphi(k)$, the total number of fractions in lowest terms is
$$
\sum_{\substack{k \mid n \\ k>1}} \varphi(k)=n-1 .
$$

The lemma is proved.
By the lemma, the number of integers in the original set is
$$
\sum_{\substack{k \mid 2015 \\ k>1}} \frac{\varphi(k)}{2}=\frac{2015-1}{2}=1007 \text {. }
$$

[Note] For problems involving the greatest common divisor or the least common multiple, setting $d=(a, b)$, and $a=A d, b=B d$ (where $A$ and $B$ are coprime integers) is a common transformation. Using this transformation can simplify the problem and pave the way for solving it.