Let $x=2^{s} x_{1}$ and $y=2^{t} y_{1}$ where $x_{1}$ and $y_{1}$ are odd integers. Without loss of generality we can assume that $s \geq t$. We have

$$z=\frac{2^{s+t+2} x_{1} y_{1}}{2^{t}\left(2^{s-t} x_{1}+y_{1}\right)}=\frac{2^{s+2} x_{1} y_{1}}{2^{s-t} x_{1}+y_{1}}$$

If $s \neq t$, then the denominator is odd and therefore $z$ is even. So we have $s=t$ and $z=2^{s+2} x_{1} y_{1} /\left(x_{1}+y_{1}\right)$. Let $x_{1}=d x_{2}, y_{1}=d y_{2}$ with $\operatorname{gcd}\left(x_{2}, y_{2}\right)=1$. So $z=$ $2^{s+2} d x_{2} y_{2} /\left(x_{2}+y_{2}\right)$. As $z$ is odd, it must be that $x_{2}+y_{2}$ is divisible by $2^{s+2} \geq 4$, so $x_{2}+y_{2}$ is divisible by 4. As $x_{2}$ and $y_{2}$ are odd integers, one of them, say $x_{2}$ is congruent to 3 modulo 4. But $\operatorname{gcd}\left(x_{2}, x_{2}+y_{2}\right)=1$, so $x_{2}$ is a divisor of $z$.

1. Given: $z = \frac{4xy}{x+y}$ is an odd integer, where $x$ and $y$ are positive integers.
2. Claim: $z$ is less than $x$ or $y$. Assume for contradiction that $z > x$ and $z > y$.
3. Contradiction: If $z > x$ and $z > y$, then:
$$\frac{4xy}{x+y} > x \quad \text{and} \quad \frac{4xy}{x+y} > y$$
This implies:
$$4xy > x(x+y) \quad \text{and} \quad 4xy > y(x+y)$$
Simplifying these inequalities:
$$4xy > x^2 + xy \quad \Rightarrow \quad 3xy > x^2 \quad \Rightarrow \quad 3y > x$$
$$4xy > y^2 + xy \quad \Rightarrow \quad 3xy > y^2 \quad \Rightarrow \quad 3x > y$$
Adding these inequalities:
$$3y + 3x > x + y \quad \Rightarrow \quad 2(x + y) > 0$$
This is always true, so the contradiction does not arise from this step. Instead, we need to consider the properties of $z$.

4. Substitution: Assume $z = y + t$ for some $t \in \mathbb{Z}_+$. Substituting into the equation:
$$z = \frac{4xy}{x+y} = y + t$$
leads to:
$$\frac{4xy}{x+y} = y + t$$
Multiplying both sides by $x + y$:
$$4xy = (y + t)(x + y)$$
Expanding and rearranging:
$$4xy = xy + y^2 + tx + ty \quad \Rightarrow \quad 3xy = y^2 + tx + ty$$
$$y^2 + (t - 3x)y + tx = 0$$

5. Discriminant Analysis: Considering the discriminant $\Delta_y$ of the quadratic equation:
$$\Delta_y = (t - 3x)^2 - 4 \cdot 1 \cdot tx = (t - 3x)^2 - 4tx$$
For $y$ to be an integer, $\Delta_y$ must be a perfect square. Let:
$$(t - 3x)^2 - 4tx = k^2$$
Rearranging:
$$(t - 3x)^2 + 4x^2 = k^2 + 4tx$$
This can be written as:
$$(2y + t - 3x)^2 + (4x)^2 = (5x - t)^2$$

6. **Existence of $r$ and $s$**: There exist $r, s \in \mathbb{Z}_+$ such that:
$$2y + t - 3x = r^2 - s^2, \quad 4x = 2rs, \quad 5x - t = r^2 + s^2$$
Solving these:
$$x = \frac{rs}{2}, \quad y = r^2 - \frac{rs}{2}, \quad z = y + t = 2rs - s^2$$
Since $z$ is odd, one of $r$ or $s$ must be even and the other odd.

7. Conclusion: If $r$ is even and $s$ is odd, then:
$$z = 2rs - s^2 \equiv 3 \pmod{4}$$
This implies that $z$ has a divisor of the form $4n - 1$.