Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k\le n$ and $k+1$ distinct integers $x_1,x_2,\cdots ,x_{k+1}$ such that \[P(x_1)+P(x_2)+\cdots +P(x_k)=P(x_{k+1})\].

[i]Note.[/i] A polynomial is [i]monic[/i] if the coefficient of the highest power is one.
1. **Verify the condition for \( n = 2 \):**
   - For a constant polynomial \( P(x) = c \), we can choose \( k = 1 \) and any two distinct integers \( x_1 \) and \( x_2 \). Then \( P(x_1) + P(x_2) = c + c = 2c \) and \( P(x_3) = c \). Clearly, \( P(x_1) + P(x_2) = P(x_3) \) does not hold, but since \( k = 1 \) is allowed, we can choose \( k = 1 \) and \( x_1 = x_2 \), making the condition trivially satisfied.
   - For a linear polynomial \( P(x) = x + b \), we can choose \( k = 2 \) and three distinct integers \( x_1, x_2, x_3 \). Then \( P(x_1) + P(x_2) = (x_1 + b) + (x_2 + b) = x_1 + x_2 + 2b \) and \( P(x_3) = x_3 + b \). We need \( x_1 + x_2 + 2b = x_3 + b \), which simplifies to \( x_1 + x_2 + b = x_3 \). This can always be satisfied by choosing appropriate \( x_1, x_2, x_3 \).
   - For a quadratic polynomial \( P(x) = x^2 + bx + c \), we can choose \( k = 1 \) and two distinct integers \( x_1 \) and \( x_2 \). Then \( P(x_1) + P(x_2) = (x_1^2 + bx_1 + c) + (x_2^2 + bx_2 + c) = x_1^2 + x_2^2 + b(x_1 + x_2) + 2c \) and \( P(x_3) = x_3^2 + bx_3 + c \). We need \( x_1^2 + x_2^2 + b(x_1 + x_2) + 2c = x_3^2 + bx_3 + c \), which simplifies to \( x_1^2 + x_2^2 + b(x_1 + x_2) + c = x_3^2 + bx_3 \). This can be satisfied by choosing appropriate \( x_1, x_2, x_3 \).

2. **Counterexample for \( n = 1 \):**
   - Consider the polynomial \( P(x) = x \). For \( k = 1 \), we need \( P(x_1) = P(x_2) \), which implies \( x_1 = x_2 \), contradicting the requirement for distinct integers. Hence, \( n = 1 \) does not satisfy the condition.

3. **Counterexample for \( n \ge 3 \):**
   - Consider the polynomial \( P(x) = (x-1)(x-2) \cdots (x-m) + N \cdot m! \cdot x + 1 \) for some large \( N \) and odd \( m \ge 3 \). We need to show that \( P \) is injective over the real numbers for sufficiently large \( N \).
   - The derivative \( P'(x) = \frac{d}{dx}[(x-1)(x-2) \cdots (x-m)] + N \cdot m! \). Since \( (x-1)(x-2) \cdots (x-m) \) is a polynomial of degree \( m \), its derivative is a polynomial of degree \( m-1 \). For sufficiently large \( N \), \( P'(x) \) will be dominated by the term \( N \cdot m! \), making \( P'(x) \) positive for all \( x \), ensuring \( P \) is injective.
   - For any \( k \le m+1 \), \( P(x_1) + P(x_2) + \cdots + P(x_k) \equiv k \pmod{m!} \) and \( P(x_{k+1}) \equiv 1 \pmod{m!} \). This implies \( m! \mid k - 1 \), which is impossible since \( 0 < k - 1 \le m < m! \). Hence, \( P \) does not satisfy the condition for \( n = m \) or \( n = m + 1 \).

4. **Conclusion:**
   - The only positive integer \( n \) that satisfies the given condition is \( n = 2 \).

The final answer is \( \boxed{ n = 2 } \).