Let $ n$ be an integer, $ n\geq 2$. Find all sets $ A$ with $ n$ integer elements such that the sum of any nonempty subset of $ A$ is not divisible by $ n\plus{}1$.
1. Let \( A = \{a_1, a_2, \ldots, a_n\} \) be a set of \( n \) integers. We need to show that all elements of \( A \) are congruent modulo \( n+1 \) and that their common residue \( p \) satisfies \( \gcd(p, n+1) = 1 \).

2. Consider the sums \( s_1 = a_1 \), \( s_2 = a_1 + a_2 \), \(\ldots\), \( s_n = a_1 + a_2 + \cdots + a_n \). None of these sums can be congruent to \( 0 \pmod{n+1} \) because if \( s_i \equiv 0 \pmod{n+1} \) for some \( i \), then taking the difference \( s_i - s_j = a_{j+1} + \cdots + a_i \equiv 0 \pmod{n+1} \) for \( i > j \) would imply that a nonempty subset of \( A \) sums to a multiple of \( n+1 \), which contradicts the problem's condition.

3. Since none of the sums \( s_1, s_2, \ldots, s_n \) are congruent to \( 0 \pmod{n+1} \), they must cover all the residues \( 1, 2, \ldots, n \) modulo \( n+1 \).

4. Now consider the sum \( a_1 + a_3 \). This sum must be congruent to one of the sums \( s_i \) modulo \( n+1 \). We analyze the possible cases:
   - If \( a_1 + a_3 \equiv s_i \pmod{n+1} \) for some \( i > 3 \), then \( s_i - (a_1 + a_3) = a_2 + a_4 + \cdots + a_i \equiv 0 \pmod{n+1} \), which is a contradiction.
   - If \( a_1 + a_3 \equiv s_3 \pmod{n+1} \), then \( a_1 + a_3 \equiv a_1 + a_2 + a_3 \pmod{n+1} \), which implies \( a_2 \equiv 0 \pmod{n+1} \), a contradiction.
   - If \( a_1 + a_3 \equiv s_1 \pmod{n+1} \), then \( a_1 + a_3 \equiv a_1 \pmod{n+1} \), which implies \( a_3 \equiv 0 \pmod{n+1} \), a contradiction.

5. The only possibility left is \( a_1 + a_3 \equiv s_2 \pmod{n+1} \), which implies \( a_1 + a_3 \equiv a_1 + a_2 \pmod{n+1} \), so \( a_2 \equiv a_3 \pmod{n+1} \).

6. By similar reasoning, we can show that all elements in the set \( A \) are congruent modulo \( n+1 \). Let \( a_1 \equiv a_2 \equiv \cdots \equiv a_n \equiv p \pmod{n+1} \).

7. For the sum of any nonempty subset of \( A \) to not be divisible by \( n+1 \), \( p \) must satisfy \( \gcd(p, n+1) = 1 \). If \( \gcd(p, n+1) \neq 1 \), then \( kp \) for some \( 0 < k < n+1 \) could be divisible by \( n+1 \), which would contradict the problem's condition.

The final answer is \( \boxed{ A = \{p, p, \ldots, p\} } \) where \( \gcd(p, n+1) = 1 \).