The real function $f$ is defined for $\forall$ $x\in \mathbb{R}$ and $f(0)=0$. Also $f(9+x)=f(9-x)$ and $f(x-10)=f(-x-10)$ for $\forall$ $x\in \mathbb{R}$. What’s the least number of zeros $f$ can have in the interval $[0;2014]$? Does this change, if $f$ is also continuous?
1. Given the function \( f \) defined for all \( x \in \mathbb{R} \) with \( f(0) = 0 \), and the properties:
   \[
   f(9 + x) = f(9 - x) \quad \text{and} \quad f(x - 10) = f(-x - 10) \quad \forall x \in \mathbb{R}
   \]
   we need to determine the least number of zeros \( f \) can have in the interval \([0, 2014]\).

2. From the first property \( f(9 + x) = f(9 - x) \), we can rewrite it as:
   \[
   f(9 + (x - 9)) = f(9 - (x - 9)) \implies f(x) = f(18 - x)
   \]
   This shows that \( f \) is symmetric about \( x = 9 \).

3. From the second property \( f(x - 10) = f(-x - 10) \), we can rewrite it as:
   \[
   f((x + 10) - 10) = f(-(x + 10) - 10) \implies f(x) = f(-x - 20)
   \]
   This shows that \( f \) is symmetric about \( x = -10 \).

4. Combining these symmetries, we get:
   \[
   f(x) = f(18 - x) = f(x - 38)
   \]
   This implies that \( f \) has a period of 38.

5. Since \( f \) is periodic with period 38, we can find the zeros of \( f \) within the interval \([0, 2014]\). Given \( f(0) = 0 \), we have:
   \[
   f(38n) = 0 \quad \text{for all integers } n
   \]
   and since \( f(x) = f(18 - x) \), we also have:
   \[
   f(38n + 18) = 0 \quad \text{for all integers } n
   \]

6. To find the number of zeros in \([0, 2014]\), we calculate:
   \[
   \text{For } f(38n) = 0: \quad 38n \leq 2014 \implies n \leq \left\lfloor \frac{2014}{38} \right\rfloor = 53
   \]
   This gives us 54 zeros (including \( n = 0 \)).

7. Similarly, for \( f(38n + 18) = 0 \):
   \[
   38n + 18 \leq 2014 \implies 38n \leq 1996 \implies n \leq \left\lfloor \frac{1996}{38} \right\rfloor = 52
   \]
   This gives us 53 zeros.

8. Therefore, the total number of zeros in \([0, 2014]\) is:
   \[
   54 + 53 = 107
   \]

9. To show that this minimum is achievable, consider the function:
   \[
   f(x) = \cos\left(\frac{\pi(x - 9)}{19}\right) - \cos\left(\frac{9\pi}{19}\right)
   \]
   This function has zeros precisely at \( x \equiv 0 \pmod{38} \) and \( x \equiv 18 \pmod{38} \).

The final answer is \(\boxed{107}\).