Let $ f(x)$ be a polynomial with the following properties:
(i)  $ f(0)\equal{}0$; (ii) $ \frac{f(a)\minus{}f(b)}{a\minus{}b}$ is an integer for any two different integers $ a$ and $ b$. Is there a polynomial which has these properties, although not all of its coefficients are integers?
1. We are given a polynomial \( f(x) \) with the properties:
   - \( f(0) = 0 \)
   - \( \frac{f(a) - f(b)}{a - b} \) is an integer for any two different integers \( a \) and \( b \).

2. We need to determine if there exists a polynomial \( f(x) \) that satisfies these properties but does not have all integer coefficients.

3. Consider the polynomial \( f(x) = \frac{x^4 + x^2}{2} \).

4. Verify the first property:
   \[
   f(0) = \frac{0^4 + 0^2}{2} = 0
   \]
   Thus, \( f(0) = 0 \).

5. Verify the second property:
   \[
   \frac{f(a) - f(b)}{a - b} = \frac{\frac{a^4 + a^2}{2} - \frac{b^4 + b^2}{2}}{a - b} = \frac{1}{2} \cdot \frac{a^4 + a^2 - b^4 - b^2}{a - b}
   \]
   Simplify the numerator:
   \[
   a^4 + a^2 - b^4 - b^2 = (a^4 - b^4) + (a^2 - b^2)
   \]
   Factorize the terms:
   \[
   a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = (a^2 + b^2)(a - b)(a + b)
   \]
   \[
   a^2 - b^2 = (a - b)(a + b)
   \]
   Combine the factorizations:
   \[
   a^4 + a^2 - b^4 - b^2 = (a - b)(a^2 + b^2)(a + b) + (a - b)(a + b)
   \]
   Factor out \( (a - b) \):
   \[
   a^4 + a^2 - b^4 - b^2 = (a - b) \left[ (a^2 + b^2)(a + b) + (a + b) \right] = (a - b)(a + b)(a^2 + b^2 + 1)
   \]
   Substitute back into the original expression:
   \[
   \frac{f(a) - f(b)}{a - b} = \frac{1}{2} \cdot (a + b)(a^2 + b^2 + 1)
   \]

6. Analyze the expression \( \frac{1}{2} (a + b)(a^2 + b^2 + 1) \):
   - If \( a \) and \( b \) have the same parity (both even or both odd), then \( a + b \) is even, making \( \frac{a + b}{2} \) an integer.
   - If \( a \) and \( b \) have opposite parity (one even, one odd), then \( a^2 + b^2 \) is odd, making \( a^2 + b^2 + 1 \) even, and \( \frac{a^2 + b^2 + 1}{2} \) an integer.

7. In both cases, \( \frac{1}{2} (a + b)(a^2 + b^2 + 1) \) is an integer, satisfying the second property.

8. The polynomial \( f(x) = \frac{x^4 + x^2}{2} \) satisfies both properties, and its coefficients are not all integers (specifically, the coefficients are \( \frac{1}{2} \) for \( x^4 \) and \( x^2 \)).

The final answer is \( \boxed{ f(x) = \frac{x^4 + x^2}{2} } \).