Find all pairs $(a,b)$ of different positive integers that satisfy the equation $W(a)=W(b)$, where $W(x)=x^{4}-3x^{3}+5x^{2}-9x$.
1. Given the function \( W(x) = x^4 - 3x^3 + 5x^2 - 9x \), we need to find all pairs \((a, b)\) of different positive integers such that \( W(a) = W(b) \).

2. Suppose \( W(a) = W(b) \). Since \( a \) and \( b \) are different, we can write:
   \[
   a^4 - 3a^3 + 5a^2 - 9a = b^4 - 3b^3 + 5b^2 - 9b
   \]
   Rearranging terms, we get:
   \[
   a^4 - b^4 - 3(a^3 - b^3) + 5(a^2 - b^2) - 9(a - b) = 0
   \]

3. Using the factorization formulas \( a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) \), \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), and \( a^2 - b^2 = (a - b)(a + b) \), we can rewrite the equation as:
   \[
   (a - b) \left[ (a^2 + b^2)(a + b) - 3(a^2 + ab + b^2) + 5(a + b) - 9 \right] = 0
   \]

4. Since \( a \neq b \), we can divide both sides by \( a - b \):
   \[
   (a^2 + b^2)(a + b) - 3(a^2 + ab + b^2) + 5(a + b) - 9 = 0
   \]

5. Simplifying the expression inside the brackets:
   \[
   (a^2 + b^2)(a + b) - 3(a^2 + ab + b^2) + 5(a + b) - 9 = 0
   \]
   \[
   a^3 + a^2b + ab^2 + b^3 - 3a^2 - 3ab - 3b^2 + 5a + 5b - 9 = 0
   \]

6. Grouping like terms:
   \[
   a^3 + b^3 + a^2b + ab^2 - 3a^2 - 3b^2 - 3ab + 5a + 5b - 9 = 0
   \]

7. Rearranging terms, we get:
   \[
   (a^2 + b^2 + 5)(a + b - 3) = 3(ab - 2)
   \]

8. Consider the case \( ab = 2 \):
   \[
   (a^2 + b^2 + 5)(a + b - 3) = 0
   \]
   Since \( a \) and \( b \) are positive integers, \( a \) and \( b \) must be 1 and 2 (or vice versa). Thus, the pairs \((a, b)\) are \((1, 2)\) and \((2, 1)\).

9. Consider the case \( ab = 1 \):
   \[
   (a^2 + b^2 + 5)(a + b - 3) = -3
   \]
   Since \( a \) and \( b \) are positive integers, the only solution is \( a = b = 1 \), which is not allowed since \( a \neq b \).

10. Now suppose \( ab > 2 \). Let \( ab = x^2 \) and \( a + b > 2 \). We have:
    \[
    3 = \frac{(a^2 + b^2 + 5)(a + b - 3)}{ab - 2} \ge \frac{(2ab + 5)(2\sqrt{ab} - 3)}{ab - 2}
    \]
    \[
    3 \ge \frac{(2x^2 + 5)(2x - 3)}{x^2 - 2}
    \]

11. Expanding and simplifying the inequality:
    \[
    3(x^2 - 2) \ge (2x^2 + 5)(2x - 3)
    \]
    \[
    3x^2 - 6 \ge 4x^3 - 9x^2 + 10x - 15
    \]
    \[
    0 \ge 4x^3 - 12x^2 + 10x - 9
    \]

12. Consider the function \( f(x) = 4x^3 - 12x^2 + 10x - 9 \). The derivative is:
    \[
    f'(x) = 12x^2 - 24x + 10
    \]
    Since \( f'(x) > 0 \) for all \( x \), \( f(x) \) is strictly increasing. Therefore, \( f(x) < 0 \) for \( x < 2 \).

13. Thus, \( ab \le 3 \). If \( ab = 3 \), then \( a = 1 \) and \( b = 3 \) (or vice versa), which does not satisfy the equation.

14. If \( x \le \frac{3}{2} \), then \( ab \le \frac{9}{4} = 2.25 \), so \( ab \le 2 \).

Conclusion:
The only pairs \((a, b)\) that satisfy the equation are \((1, 2)\) and \((2, 1)\).

The final answer is \( \boxed{ (1, 2) } \) and \((2, 1)\).