To solve this problem, we need to analyze the given system of equations:

$$\begin{align*} 1) \quad & x + y = z + u,\\ 2) \quad & 2xy = zu. \end{align*}$$

Our goal is to find the greatest value of the real constant $m$ such that $m \leq \frac{x}{y}$ for any positive integer solution $(x, y, z, u)$ with $x \geq y$.

### Step 1: Express $z$ and $u$ in terms of $x$ and $y$

From equation (1), we have:

$$z + u = x + y.$$

Using equation (2):

$$zu = 2xy.$$

These two equations describe a pair of numbers $z$ and $u$ which, together, sum to $x + y$ and have a product of $2xy$.

### Step 2: Solve the quadratic equation

Consider $z$ and $u$ as the roots of the quadratic equation:

$$t^2 - (x+y)t + 2xy = 0.$$

Using the quadratic formula:

$$t = \frac{(x+y) \pm \sqrt{(x+y)^2 - 8xy}}{2}.$$

The discriminant of the quadratic must be non-negative for real solutions $z$ and $u$, so:

$$(x+y)^2 - 8xy \geq 0.$$

This simplifies to:

$$x^2 + 2xy + y^2 - 8xy \geq 0,$$

or

$$x^2 - 6xy + y^2 \geq 0.$$

### Step 3: Transform the inequality

Rearrange the terms:

$$(x-y)^2 \geq 4xy.$$

Dividing throughout by $y^2$ (assuming $y > 0$), we get:

$$\left( \frac{x}{y} - 1 \right)^2 \geq 4 \cdot \frac{x}{y}.$$

Let $\frac{x}{y} = k$ where $k \geq 1$. This gives:

$$(k - 1)^2 \geq 4k.$$

Expanding and rearranging:

$$k^2 - 6k + 1 \geq 0.$$

We solve the quadratic inequality using the quadratic formula:

$$k = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}.$$

Since $k = \frac{x}{y} \geq 1$, we take the larger root, giving us:

$$k \geq 3 + 2\sqrt{2}.$$

Thus, the greatest value of $m$ is:

$$\boxed{3 + 2\sqrt{2}}.$$