Find all sets of $n\ge 2$ consecutive integers $\{a+1,a+2,...,a+n\}$  where  $a\in Z$, in which one of the numbers is equal to the sum of all the others.

(Bogdan Rublev)
1. **Initial Considerations:**
   - For \( n = 2 \), no two consecutive integers can satisfy the condition since no two consecutive integers are equal.
   - Therefore, we consider \( n \geq 3 \).
   - If \(\{a+1, a+2, \ldots, a+n\}\) works, then \(\{-a-1, -a-2, \ldots, -a-n\}\) also works due to symmetry.

2. **Nonnegative Integers Case:**
   - Assume the set \(\{a+1, a+2, \ldots, a+n\}\) consists of nonnegative integers.
   - The greatest element must be greater than or equal to the sum of all other elements.
   - The sum of the set is:
     \[
     S = (a+1) + (a+2) + \cdots + (a+n) = na + \frac{n(n+1)}{2}
     \]
   - If one element, say \(a+k\), is equal to the sum of the others:
     \[
     a+k = S - (a+k)
     \]
     \[
     2(a+k) = S
     \]
     \[
     2(a+k) = na + \frac{n(n+1)}{2}
     \]
     \[
     2a + 2k = na + \frac{n(n+1)}{2}
     \]
     \[
     2k = (n-2)a + \frac{n(n+1)}{2}
     \]
   - For \(a+n\) to be the greatest element:
     \[
     a+n \geq a+(n-1)+a+(n-2) \implies a+n \leq 3
     \]
   - Given \(a \geq -1\) and \(n \geq 3\), possible pairs \((a, n)\) are:
     \[
     (a, n) = (-1, 3), (-1, 4), (0, 3)
     \]
   - Checking these pairs:
     - \((a, n) = (-1, 3)\) gives \(\{0, 1, 2\}\), which does not work.
     - \((a, n) = (-1, 4)\) gives \(\{0, 1, 2, 3\}\), which works.
     - \((a, n) = (0, 3)\) gives \(\{1, 2, 3\}\), which works.

3. **Negative Integers Case:**
   - Similarly, \(\{-3, -2, -1, 0\}\) and \(\{-3, -2, -1\}\) work.

4. **Mixed Positive and Negative Integers:**
   - If the set contains both positive and negative integers:
     - If the set has the same number of positive and negative integers, then \(0\) is equal to the sum of all other elements:
       \[
       \{-k, -k+1, \ldots, k-1, k\}
       \]
     - If there are more positive integers than negative integers:
       \[
       \{-n, -n+1, \ldots, n+m-1, n+m\}
       \]
       - For \(m=1\), the sum of all elements is \(n+1\), which must be even, so \(n\) must be odd:
         \[
         \{-2k+1, -2k+2, \ldots, 2k\}
         \]
       - For \(m=2\), the sum of all elements is \(2n+3\), which is odd, a contradiction.
       - For \(m \geq 3\), since the greatest element must be greater than or equal to all other elements:
         \[
         m+n \geq m+n-2+m+n-1 \implies m+n \leq 3
         \]
         - This is a contradiction since \(n \geq 1\).

5. **Conclusion:**
   - The sets that satisfy the property are:
     \[
     \{0, 1, 2, 3\}, \{1, 2, 3\}, \{-3, -2, -1, 0\}, \{-3, -2, -1\}, \{-k, -k+1, \ldots, k-1, k\}, \{-2k+1, -2k+2, \ldots, 2k\}, \{-2k, -2k+1, \ldots, 2k+1\}
     \]

The final answer is \( \boxed{ \{0, 1, 2, 3\}, \{1, 2, 3\}, \{-3, -2, -1, 0\}, \{-3, -2, -1\}, \{-k, -k+1, \ldots, k-1, k\}, \{-2k+1, -2k+2, \ldots, 2k\}, \{-2k, -2k+1, \ldots, 2k+1\} } \).