Example 8 Find all positive integers $k$ that satisfy the following proposition: If $F(x)$ is a polynomial with integer coefficients, and satisfies the condition: for any $c \in\{0,1, \cdots, k+1\}, 0 \leqslant F(c) \leqslant k$, then $F(0)=F(1)=\cdots=F(k+1)$.
From the given conditions, it is easy to see that $F(k+1)=F(0)$, hence $F(x)-F(0)=x(x-k-1) G(x)$, where $G(x)$ is a polynomial with integer coefficients. Further, by the factor theorem, when $n \geqslant 3$, we have
$$
G(x)=x(x-2)(x-3) \cdots(x-k+1)(x-k-k) H(x),
$$

where $H(x)$ is a polynomial with integer coefficients. Finally, when $n \geqslant 4$, it is easy to prove that $H(1)=H(k)=0$. Thus, the proposition holds for $k \geqslant 4$. For $k=1,2,3$, counterexamples can be easily found.

Solution: This proposition holds if and only if $k \geqslant 4$.
We first prove that the proposition holds for any $k \geqslant 4$.
For an integer-coefficient polynomial $F(x)$ satisfying the given conditions, since $F(k+1)-F(0)$ is a multiple of $k+1$, and $|F(k+1)-F(0)| \leqslant k$, it follows that $F(k+1)=F(0)$. Therefore, $F(x)-F(0)=x(x-k-1) G(x)$, where $G(x)$ is a polynomial with integer coefficients. Thus, for any $c \in\{1,2, \cdots, k\}$, we have
$$
k \geqslant|F(c)-F(0)|=c(k+1-c)|G(c)| .
$$

Notice that the inequality $c(k+1-c)>k$, i.e., $(c-1)(k-c)>0$ holds for any $c \in\{2,3, \cdots, k-1\}$, and when $c \geqslant 3$, the set $\{2,3, \cdots, k-1\}$ is non-empty. Therefore, for any $c$ in this set, from (1) we get $|G(c)|<1$, hence $|H(c)|<1$, so $H(c)=0$.
Thus, we have proven that the proposition holds for any $k \geqslant 4$.
The above proof also provides clues for listing counterexamples when $k=1,2,3$. These counterexamples are:
$$
\begin{array}{l}
\text { When } k=1 \text {, } F(x)=x(2-x) \text {; } \\
\text { When } k=2 \text {, } F(x)=x(3-x) \text {; } \\
\text { When } k=3 \text {, } F(x)=x(4-x)(x-2)^{2} \text {. }
\end{array}
$$