We say that a quadruple of nonnegative real numbers $(a,b,c,d)$ is [i]balanced [/i]if $$a+b+c+d=a^2+b^2+c^2+d^2.$$
    Find all positive real numbers $x$ such that $$(x-a)(x-b)(x-c)(x-d)\geq 0$$ for every balanced quadruple $(a,b,c,d)$. \\ \\
(Ivan Novak)
To solve the problem, we need to find all positive real numbers \( x \) such that the inequality 
\[ (x-a)(x-b)(x-c)(x-d) \geq 0 \]
holds for every balanced quadruple \((a, b, c, d)\) where 
\[ a + b + c + d = a^2 + b^2 + c^2 + d^2. \]

1. **Initial Case Analysis:**
   - Consider the balanced quadruple \((a, b, c, d) = (1, 0, 0, 0)\):
     \[
     a + b + c + d = 1 + 0 + 0 + 0 = 1
     \]
     \[
     a^2 + b^2 + c^2 + d^2 = 1^2 + 0^2 + 0^2 + 0^2 = 1
     \]
     This satisfies the balanced condition. The inequality becomes:
     \[
     (x-1)x^3 \geq 0
     \]
     This implies:
     \[
     x \geq 1
     \]

2. **Second Case Analysis:**
   - Consider the balanced quadruple \((a, b, c, d) = \left(\frac{3}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\):
     \[
     a + b + c + d = \frac{3}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 3
     \]
     \[
     a^2 + b^2 + c^2 + d^2 = \left(\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{9}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 3
     \]
     This satisfies the balanced condition. The inequality becomes:
     \[
     \left(x - \frac{3}{2}\right)\left(x - \frac{1}{2}\right)^3 \geq 0
     \]
     Since \( x \geq 1 > \frac{1}{2} \), we get:
     \[
     x \geq \frac{3}{2}
     \]

3. **General Analysis:**
   - We need to ensure that \( x \geq a, b, c, d \) for all balanced quadruples. Consider the sum of squares condition:
     \[
     a^2 + b^2 + c^2 + d^2 = a + b + c + d
     \]
     Using the Cauchy-Schwarz inequality:
     \[
     (a^2 + b^2 + c^2 + d^2)(1^2 + 1^2 + 1^2 + 1^2) \geq (a + b + c + d)^2
     \]
     \[
     4(a^2 + b^2 + c^2 + d^2) \geq (a + b + c + d)^2
     \]
     Given \( a^2 + b^2 + c^2 + d^2 = a + b + c + d \), we have:
     \[
     4(a + b + c + d) \geq (a + b + c + d)^2
     \]
     Let \( S = a + b + c + d \), then:
     \[
     4S \geq S^2 \implies S^2 - 4S \leq 0 \implies S(S - 4) \leq 0
     \]
     Since \( S \geq 0 \), we get:
     \[
     0 \leq S \leq 4
     \]
     The maximum value of any \( a, b, c, d \) is \(\frac{3}{2}\) because:
     \[
     \sum (2a - 1)^2 = 4 \implies a, b, c, d \leq \frac{3}{2}
     \]
     Therefore, \( x \geq \frac{3}{2} \) works for all balanced quadruples.

Conclusion:
The answer is all real \( x \geq \frac{3}{2} \).

The final answer is \( \boxed{ x \geq \frac{3}{2} } \).