We are asked to solve the equation $n+ \lfloor \sqrt{n} \rfloor + \lfloor \sqrt[3]{n} \rfloor = 2014$ for positive real numbers $n$.

To begin, we denote:
- $x = \lfloor \sqrt{n} \rfloor$,
- $y = \lfloor \sqrt[3]{n} \rfloor$.

Thus, we have:
$$x \leq \sqrt{n} < x+1$$
$$y \leq \sqrt[3]{n} < y+1$$

Therefore,
$$x^2 \leq n < (x+1)^2$$
$$y^3 \leq n < (y+1)^3$$

Substituting these into the original equation, we have:
$$n + x + y = 2014$$

Next, we will approximate values to find the range for $n$.

1. **Approximate $\sqrt{n}$ and $\sqrt[3]{n}$**

Since $x = \lfloor \sqrt{n} \rfloor$ and $y = \lfloor \sqrt[3]{n} \rfloor$, assume that:
$$n \approx x^2 \quad \text{and} \quad n \approx y^3$$

2. **Estimate $n$:**

From the equation, $n + x + y = 2014$, express $x$ and $y$ in terms of $n$:
$$x \approx \sqrt{n} \quad \text{and} \quad y \approx \sqrt[3]{n}$$

Substituting:
$$n + \sqrt{n} + \sqrt[3]{n} \approx 2014$$

Let's try some approximate calculations:

3. **Trial Calculation for $n$:**

Suppose $n = 1958$, then:
$$\sqrt{1958} \approx 44.23 \quad \Rightarrow \quad \lfloor \sqrt{1958} \rfloor = 44$$
$$\sqrt[3]{1958} \approx 12.57 \quad \Rightarrow \quad \lfloor \sqrt[3]{1958} \rfloor = 12$$

Hence,
$$1958 + 44 + 12 = 2014$$

This satisfies the equation.

Thus, the solution to the equation in positive real numbers is:
$$\boxed{1958}$$