Find all positive integers $k$ for which the following assertion holds:
If $F(x)$ is polynomial with integer coefficients ehich satisfies $F(c) \leq k$ for all $c \in \{0,1, \cdots,k+1 \}$, then \[F(0)= F(1) = \cdots =F(k+1).\]
1. **Understanding the problem**: We need to find all positive integers \( k \) such that if a polynomial \( F(x) \) with integer coefficients satisfies \( 0 \leq F(c) \leq k \) for all \( c \in \{0, 1, \ldots, k+1\} \), then \( F(0) = F(1) = \cdots = F(k+1) \).

2. **Analyzing the polynomial**: Let \( F(x) \) be a polynomial with integer coefficients. We are given that \( 0 \leq F(c) \leq k \) for all \( c \in \{0, 1, \ldots, k+1\} \). This means that \( F(x) \) takes on integer values within the range \([0, k]\) at these points.

3. **Considering the degree of the polynomial**: Suppose \( F(x) \) is a polynomial of degree \( n \). If \( n \geq k+1 \), then \( F(x) \) can have at most \( n \) distinct roots. However, if \( F(x) \) is not constant, it can take on different values at different points.

4. **Using the given condition**: Since \( F(x) \) is bounded by \( k \) at \( k+2 \) points, we need to check if this forces \( F(x) \) to be constant. If \( F(x) \) is not constant, it must have at least \( k+1 \) distinct values within the range \([0, k]\).

5. **Checking for small values of \( k \)**:
   - For \( k = 1 \): The polynomial \( F(x) \) must satisfy \( 0 \leq F(c) \leq 1 \) for \( c \in \{0, 1, 2\} \). If \( F(x) \) is not constant, it must take on at least two distinct values within \([0, 1]\). However, since there are only two possible values (0 and 1), \( F(x) \) must be constant to satisfy the condition.
   - For \( k = 2 \): The polynomial \( F(x) \) must satisfy \( 0 \leq F(c) \leq 2 \) for \( c \in \{0, 1, 2, 3\} \). If \( F(x) \) is not constant, it must take on at least three distinct values within \([0, 2]\). However, since there are only three possible values (0, 1, and 2), \( F(x) \) must be constant to satisfy the condition.
   - For \( k = 3 \): The polynomial \( F(x) \) must satisfy \( 0 \leq F(c) \leq 3 \) for \( c \in \{0, 1, 2, 3, 4\} \). If \( F(x) \) is not constant, it must take on at least four distinct values within \([0, 3]\). However, since there are only four possible values (0, 1, 2, and 3), \( F(x) \) must be constant to satisfy the condition.

6. **Generalizing the result**: For \( k \geq 4 \), the polynomial \( F(x) \) must satisfy \( 0 \leq F(c) \leq k \) for \( c \in \{0, 1, \ldots, k+1\} \). If \( F(x) \) is not constant, it must take on at least \( k+1 \) distinct values within \([0, k]\). However, since there are only \( k+1 \) possible values, \( F(x) \) must be constant to satisfy the condition.

7. **Conclusion**: The assertion holds for all positive integers \( k \).

The final answer is \( \boxed{ k \geq 1 } \).