We are given the equation:

$$\left(1+ \frac{x}{y+z} \right)^2 + \left(1+ \frac{y}{z+x} \right)^2 + \left(1+ \frac{z}{x+y} \right)^2 = \frac{27}{4}$$

and need to solve for $x, y, z \in \mathbb{N}$.

Let's start by expanding each term inside the equation:

$$\left(1+ \frac{x}{y+z} \right)^2 = 1 + 2\frac{x}{y+z} + \frac{x^2}{(y+z)^2}$$

Similarly, we have:

$$\left(1+ \frac{y}{z+x} \right)^2 = 1 + 2\frac{y}{z+x} + \frac{y^2}{(z+x)^2}$$

$$\left(1+ \frac{z}{x+y} \right)^2 = 1 + 2\frac{z}{x+y} + \frac{z^2}{(x+y)^2}$$

Substituting these expanded forms back into the original equation gives:

$$1 + 2\frac{x}{y+z} + \frac{x^2}{(y+z)^2} + 1 + 2\frac{y}{z+x} + \frac{y^2}{(z+x)^2} + 1 + 2\frac{z}{x+y} + \frac{z^2}{(x+y)^2} = \frac{27}{4}$$

Combine the constant terms:

$$3 + 2\left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y}\right) + \left(\frac{x^2}{(y+z)^2} + \frac{y^2}{(z+x)^2} + \frac{z^2}{(x+y)^2}\right) = \frac{27}{4}$$

Subtract 3 from both sides:

$$2\left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y}\right) + \left(\frac{x^2}{(y+z)^2} + \frac{y^2}{(z+x)^2} + \frac{z^2}{(x+y)^2}\right) = \frac{15}{4}$$

Assume $x = y = z$. Then:

$$\frac{x}{y+z} = \frac{x}{2x} = \frac{1}{2}$$

$$\frac{y}{z+x} = \frac{y}{2y} = \frac{1}{2}$$

$$\frac{z}{x+y} = \frac{z}{2z} = \frac{1}{2}$$

Thus, the equation becomes:

$$2\left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right) + \left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\right) = \frac{15}{4}$$

This simplifies to:

$$2 \times \frac{3}{2} + \frac{3}{4} = 3 + \frac{3}{4} = \frac{12}{4} + \frac{3}{4} = \frac{15}{4}$$

Thus, the condition holds true when $x = y = z$. Therefore, the solution is:

$$\boxed{x = y = z}$$