Find all pairs of positive integers $(x, y)$ such that $x<y, x$ and $y$ are coprime and $2000\left(\frac{x}{y}+\frac{y}{x}\right)$ is an odd integer.
We have $2000=16 \times 125=2^{4} \times 5^{3}$,

$$
N=2000\left(\frac{x}{y}+\frac{y}{x}\right)=16 \times 125\left(\frac{x^{2}+y^{2}}{x y}\right)=\frac{16 \times 125}{x y} \times\left(x^{2}+y^{2}\right)
$$

is an odd integer and $x16$, so $x$ is at least $16 \times 5=80$, since $x$ divides 2000; as also $x y$ divides 2000, it would result that $y \leq 25<80=x$, which is not allowed. Therefore, $x=16$ and, since $x$ and $y$ are coprime and $y$ divides 2000, necessarily $y=25$ or 125.

$2^{2}$ Case: If 16 divides $y$, one possibility is $y=16$, when $x$ can only be 1 or 5, since $x<y$ and $x y$ divides 2000. The other options for $y$ are $16 \times 5, 16 \times 25$ or 2000, when the only option for $x$ is, always, 1.

Thus, there are seven solutions, namely,

$$
(16,25),(16,125),(1,16),(5,16),(1,80),(1,400) \text { and }(1,2000)
$$

Solution 2: Since $N$ is an odd integer, it follows that $\left(2^{4} \times 5^{3}\right) /(x y)$ and $x^{2}+y^{2}$ are odd integers. The options for $x y$ are $2^{4}, 2^{4} \times 5, 2^{4} \times 5^{2}$, and $2^{4} \times 5^{3}$. Moreover, $x$ and $y$ must have different parities to ensure that $x^{2}+y^{2}$ is odd. Let's determine $x$ and $y$ for each of these options, remembering that $x<y$.

| $x y$ | $x$ | $y$ |
| :---: | :---: | :---: |
| $2^{4}$ | 1 | $2^{4}$ |
| $2^{4} \times 5$ | 1 | $2^{4} \times 5$ |
|  | 5 | $2^{4}$ |
| $2^{4} \times 5^{2}$ | 1 | $2^{4} \times 5^{2}$ |
|  | $2^{4}$ | $5^{2}$ |
| $2^{4} \times 5^{3}$ | 1 | $2^{4} \times 5^{3}$ |
|  | $2^{4}$ | $5^{3}$ |

Thus, there are seven solutions, namely,

$$
(1,16),(1,80),(5,16),(1,400),(16,25),(1,2000) \text { and }(16,125)
$$