Suppose that there are rational numbers $x$ and $y$ such that $x - \frac{1}{x} + y - \frac{1}{y} = 4$.
Since $\left(-\frac{1}{x}, y\right)$, $\left(x, -\frac{1}{y}\right)$ and $\left(-\frac{1}{x}, -\frac{1}{y}\right)$ are solutions of the equation, we can assume that $x > 0$ and $y > 0$.

Letting $u = xy$, we get $x + y = \frac{4xy}{xy - 1} = \frac{4u}{u - 1}$. Note that $u > 0$ and $u \ne 1$.
Now the quadratic equation
$$T^2 - \frac{4u}{u-1}T + u = 0$$
has rational solutions $x$ and $y$. So the discriminant $\left(\frac{4u}{u-1}\right)^2 - 4u$ is a square of a rational number. Then
$$4u^2 - u(u-1)^2 = u(6u - u^2 - 1)$$
is a square of a rational number.
Let $u = \frac{q}{p}$ with distinct positive integers $p$ and $q$ such that $(p, q) = 1$. From
$$u(6u - u^2 - 1) = \frac{q(6pq - p^2 - q^2)}{p^3} = \frac{1}{p^4} \cdot pq(6pq - p^2 - q^2),$$
we know $pq(6pq - p^2 - q^2)$ is a square of an integer. Because $p, q$ and $6pq - p^2 - q^2$ are pairwise relatively prime, there are positive integers $s, t$ and $w$ such that
$$p = s^2, \quad q = t^2, \quad \text{and} \quad w^2 = 6pq - p^2 - q^2.$$
Thus $w^2 = 6s^2t^2 - s^4 - t^4 = (2st)^2 - (s^2 - t^2)^2$. Note that $(s, t) = 1$ and $s \neq t$.
Without loss of generality assume $s > t > 0$. Assume that $(s, t, w)$ is the positive integer solution of $w^2 = 6s^2t^2 - s^4 - t^4 = (2st)^2 - (s^2 - t^2)^2$ which makes $s+t$ minimum.
Because $2st$ is even in the Pythagorean triple $\{w, s^2 - t^2, 2st\}$, we know $s^2 - t^2$ is even and $s$ and $t$ are odd. So $\left(\frac{s^2 - t^2}{2}, st\right) = 1$. Then
$$\left(\frac{w}{2}\right)^2 + \left(\frac{s^2 - t^2}{2}\right)^2 = (st)^2$$
gives a primitive Pythagorean triple $\left(\frac{w}{2}, \frac{s^2 - t^2}{2}, st\right)$. So there are positive integers $m$ and $n$ such that $(m, n) = 1$, $s^2 - t^2 = 4mn$ and $st = m^2 + n^2$. We can assume that $m$ is even and $n$ is odd.
From the equality $s^2 - t^2 = (s+t)(s-t) = 4mn$, there are positive integers $A, B, C, D$ which are pairwise relatively prime satisfying
$$s + t = 2AB, \quad s - t = 2CD, \quad m = AC, \quad n = BD.$$
By plugging these to $st = m^2 + n^2$, we have $2A^2B^2 = (A^2 + D^2)(B^2 + C^2)$. Then we know $C$ is even from the assumption that $m$ is even. Note that $A, B, D$ are odd.
Now from the equality $2A^2B^2 = (A^2 + D^2)(B^2 + C^2)$, we get $A^2 = B^2 + C^2$, $2B^2 = A^2 + D^2$. Then from $A^2 = B^2 + C^2$, we get $A = a^2 + b^2$, $B = a^2 - b^2$, $C = 2ab$ with

positive integers $a$ and $b$ such that $(a,b) = 1$ and $a > b > 0$. Then $2B^2 = A^2 + D^2$
becomes $a^4 + b^4 - 6a^2b^2 = D^2$. Now $(2D)^2 = 6(a+b)^2(a-b)^2 - (a+b)^4 - (a-b)^4$.
We have a contradiction from the minimality of $s+t$ since $s+t = 2AB = 2B(a^2 + b^2) > 2a = (a+b) + (a-b) > 0$.
Therefore, there are no rational numbers $x$ and $y$ such that $x - \frac{1}{x} + y - \frac{1}{y} = 4$.