How many pairs of positive integers $(x, y)$ are there such that

$$
\frac{x y}{x+y}=144 ?
$$
The given equation is equivalent to $x y=144(x+y)=144 x+144 y$, so isolating $x$, we get $x=\frac{144 y}{y-144}$. Since $x$ and $y$ must be positive integers, the denominator $y-144$ must be a positive integer, say, $y-144=n$. Substituting this expression into the value of $x$, we get

$$
x=\frac{144(n+144)}{n}=144+\frac{144^{2}}{n}
$$

Since $x$ must be an integer, $n$ must be a divisor of $144^{2}$. Given that $144^{2}=12^{4}=2^{8} \cdot 3^{4}$, its divisors are numbers $d$ of the form $d=2^{a} \cdot 3^{b}$, with $0 \leq a \leq 8$ and $0 \leq b \leq 4$. Since there are 9 possible values for $a$ and 5 possible values for $b$, we conclude that $144^{2}$ has $9 \times 5=45$ divisors.

Thus, for each divisor $n$ of $144^{2}$, we obtain a solution

$$
(x, y)=\left(144+\frac{144^{2}}{n}, n+144\right)
$$

to the given equation $\frac{x y}{x+y}=144$. Therefore, this equation has 45 pairs of positive integers $(x, y)$ that satisfy it.