Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)\geq 0\ \forall \ x\in \mathbb{R}$, $f'(x)$ exists $\forall \ x\in \mathbb{R}$ and $f'(x)\geq 0\ \forall \ x\in \mathbb{R}$ and $f(n)=0\ \forall \ n\in \mathbb{Z}$
1. Given the function \( f: \mathbb{R} \to \mathbb{R} \) such that \( f(x) \geq 0 \) for all \( x \in \mathbb{R} \), \( f'(x) \) exists for all \( x \in \mathbb{R} \), and \( f'(x) \geq 0 \) for all \( x \in \mathbb{R} \). Additionally, \( f(n) = 0 \) for all \( n \in \mathbb{Z} \).

2. Since \( f'(x) \geq 0 \) for all \( x \in \mathbb{R} \), the function \( f(x) \) is non-decreasing. This means that for any \( x_1 < x_2 \), we have \( f(x_1) \leq f(x_2) \).

3. Suppose there exists some \( a \in \mathbb{R} \) such that \( f(a) = h > 0 \). Since \( f \) is non-decreasing, for any \( x \geq a \), \( f(x) \geq h \).

4. However, we are given that \( f(n) = 0 \) for all \( n \in \mathbb{Z} \). This implies that for any integer \( n \), \( f(n) = 0 \).

5. Consider \( n \in \mathbb{Z} \) such that \( n > a \). Since \( f \) is non-decreasing and \( f(n) = 0 \), we must have \( f(x) \leq 0 \) for all \( x \geq n \).

6. Combining the facts that \( f(x) \geq 0 \) and \( f(x) \leq 0 \) for all \( x \geq n \), we conclude that \( f(x) = 0 \) for all \( x \geq n \).

7. Since \( f \) is non-decreasing and \( f(x) = 0 \) for all \( x \geq n \), it follows that \( f(x) = 0 \) for all \( x \leq n \) as well.

8. Therefore, \( f(x) = 0 \) for all \( x \in \mathbb{R} \).

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = 0 \quad \forall x \in \mathbb{R} } \)