For a polynomials $ P\in \mathbb{R}[x]$, denote $f(P)=n$ if $n$ is the smallest positive integer for which is valid
$$(\forall x\in \mathbb{R})(\underbrace{P(P(\ldots P}_{n}(x))\ldots )>0),$$
and $f(P)=0$ if such n doeas not exist. Exists polyomial $P\in \mathbb{R}[x]$ of degree $2014^{2015}$ such that  $f(P)=2015$?

(Serbia)
1. **Define the polynomial \( P_\varepsilon(x) \):**
   Consider the polynomial \( P_\varepsilon(x) = x^{2n} + x + \varepsilon \), where \( \varepsilon \) is a small positive real number and \( n \) is a positive integer.

2. **Behavior of \( P_\varepsilon(x) \) for large \( x \):**
   For large values of \( x \), the term \( x^{2n} \) dominates, making \( P_\varepsilon(x) \) positive. Specifically, \( P_\varepsilon(x) \to \infty \) as \( x \to \infty \).

3. **Behavior of \( P_\varepsilon(x) \) for small \( x \):**
   For small values of \( x \), particularly for \( x \leq 0 \), the term \( x^{2n} \) is non-negative, and the linear term \( x \) and constant term \( \varepsilon \) ensure that \( P_\varepsilon(x) \) is positive for sufficiently small \( \varepsilon \).

4. **Fixed point analysis:**
   When \( \varepsilon = 0 \), the polynomial \( P_0(x) = x^{2n} + x \) has a fixed point at \( x = 0 \). This fixed point is non-positive and does not grow beyond 0 under iteration.

5. **Continuity and the function \( g(\varepsilon) \):**
   Define \( g(\varepsilon) = f(P_\varepsilon) \), where \( f(P) \) is the smallest positive integer \( n \) such that \( P(P(\ldots P(x)\ldots)) > 0 \) for all \( x \in \mathbb{R} \). The function \( g(\varepsilon) \) is "continuous" in the sense that if \( x < y \) and \( g(x) = m > n = g(y) \), then there exists \( x < z < y \) such that \( m > g(z) > n \).

6. **Behavior of \( g(\varepsilon) \):**
   - For \( \varepsilon = 1 \), \( g(1) = 1 \) because the minimum of \( x^{2n} + x \) is strictly greater than \(-1\).
   - As \( \varepsilon \to 0 \), \( g(\varepsilon) \to +\infty \) because \( P_0(x) \) has a fixed point at 0, which never gets greater than 0.

7. **Intermediate Value Theorem (IVT) application:**
   By the Intermediate Value Theorem (IVT) for this kind of "continuous" function, for each positive integer \( k \), there exists an \( \varepsilon \) such that \( f(P_\varepsilon) = k \).

8. **Specific case for \( n = 2014^{2015} \) and \( k = 2015 \):**
   Let \( 2n = 2014^{2015} \). By the IVT, there exists an \( \varepsilon \) such that \( f(P_\varepsilon) = 2015 \).

Conclusion:
\[
\boxed{\text{True}}
\]