Example 6 Let $a, b$ be positive integers, and satisfy $2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{15}{b}}\right)$ is an integer. Then the number of such ordered pairs $(a, b)$ is $\qquad$ pairs.
$(2009$, National Junior High School Mathematics League)
First, guess that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
The following is an attempt to prove this.
Let $\frac{15}{a}=A, \frac{15}{b}=B$. According to the problem, we can assume
$\sqrt{A}+\sqrt{B}=C\left(A, B, C \in \mathbf{Q}_{+}\right)$.
Thus, $\sqrt{A}=C-\sqrt{B} \Rightarrow A=(C-\sqrt{B})^{2}$
$\Rightarrow A=C^{2}-2 C \sqrt{B}+B \Rightarrow \sqrt{B}=\frac{C^{2}+B-A}{2 C}$
$\Rightarrow \sqrt{B}$ is a rational number.
Similarly, $\sqrt{A}$ is a rational number.
Therefore, $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
Further conjecture: When $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are expressed in their simplest fractional form, the numerator is 1.
The following is an attempt to prove this.
Let $\sqrt{\frac{15}{a}}=\frac{q}{p}\left(p, q \in \mathbf{Z}_{+}\right.$, and $p$ and $q$ are coprime $)$.
Then $\left.\frac{15}{a}=\frac{q^{2}}{p^{2}} \Rightarrow a q^{2}=15 p^{2} \Rightarrow q^{2} \right\rvert\, 15 p^{2}$.
Since $p$ and $q$ are coprime, $p^{2}$ and $q^{2}$ are coprime. Hence
$q^{2} \left\lvert\, 15 \Rightarrow q^{2}=1 \Rightarrow q=1 \Rightarrow \sqrt{\frac{15}{a}}=\frac{1}{p}\right.$.
Similarly, when $\sqrt{\frac{15}{b}}$ is expressed in its simplest fractional form, the numerator is 1.
From the previous analysis, we can assume
$\sqrt{\frac{15}{a}}=\frac{1}{x}, \sqrt{\frac{15}{b}}=\frac{1}{y}\left(x, y \in \mathbf{Z}_{+}\right)$. Then $a=15 x^{2}, b=15 y^{2}$.
According to the problem, we can assume $\frac{1}{x}+\frac{1}{y}=\frac{k}{2}\left(k \in \mathbf{Z}_{+}\right)$.
Since $x \geqslant 1, y \geqslant 1$, we have
$$
\frac{k}{2}=\frac{1}{x}+\frac{1}{y} \leqslant 1+1=2 \Rightarrow k \leqslant 4 \text {. }
$$

Therefore, $k$ can only be $1,2,3,4$.
(1) When $k=1$,
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$
$\Rightarrow(x-2)(y-2)=4$
$\Rightarrow(x, y)=(4,4)$ or $(3,6)$ or $(6,3)$;
(2) When $k=2$, $(x, y)=(2,2)$;
(3) When $k=3$, $(x, y)=(2,1)$ or $(1,2)$;
(4) When $k=4$, $(x, y)=(1,1)$.

Therefore, there are 7 such ordered pairs $(a, b)$: $(240,240),(135,540),(540,135)$, $(60,60),(60,15),(15,60),(15,15)$.