Find all nonempty finite sets $X$ of real numbers such that for all $x\in X$, $x+|x| \in X$.
1. **Consider the condition given in the problem:**
   For all \( x \in X \), \( x + |x| \in X \).

2. **Analyze the case when \( x \) is positive:**
   - If \( x > 0 \), then \( |x| = x \).
   - Therefore, \( x + |x| = x + x = 2x \).
   - If \( x \in X \), then \( 2x \in X \).
   - Repeating this process, we get \( 2^n x \in X \) for all \( n \in \mathbb{N} \).
   - Since \( x > 0 \), \( 2^n x \) will grow without bound as \( n \) increases.
   - Hence, \( X \) would be infinite if it contains any positive number.

3. **Analyze the case when \( x \) is non-positive:**
   - If \( x \leq 0 \), then \( |x| = -x \).
   - Therefore, \( x + |x| = x - x = 0 \).
   - If \( x \in X \) and \( x \leq 0 \), then \( 0 \in X \).

4. **Conclusion:**
   - If \( X \) contains any positive number, \( X \) must be infinite.
   - If \( X \) is finite, it cannot contain any positive number.
   - Therefore, \( X \) must consist of non-positive numbers.
   - Since \( 0 \in X \) if \( X \) contains any non-positive number, \( X \) must be of the form \( F \cup \{0\} \), where \( F \) is a finite set of negative real numbers (or \( F \) can be empty).

The final answer is \( \boxed{ X = F \cup \{0\} } \), where \( F \) is any finite (possibly empty) set of negative real numbers.