Let $k$ be a given positive integer. Try to find all real numbers $a$, such that the quadratic equation $a x^{2}+a x+a+2006=0$ has two integer roots, and the absolute value of the difference between the two roots is not less than $2 k-1$.

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Solution: Let the two integer roots of $a x^{2}+a x+a+2006=0$ be $x_{1}$ and $x_{2}$. Then
$$
a x_{1}^{2}+a x_{1}+a+2006=0.
$$

Thus, $a=-\frac{2006}{x_{1}^{2}+x_{1}+1}=-\frac{2006}{t^{2}+t+1}(t \in \mathbf{N})$.
Conversely, if $a=-\frac{2006}{t^{2}+t+1}(t$ is any integer), then the original equation can be rewritten as
$$
-\frac{2006}{t^{2}+t+1}\left(x^{2}+x+1\right)+2006=0,
$$

which simplifies to $-x^{2}-x-1+t^{2}+t+1=0$.
Rearranging and factoring, we get
$$
(x-t)(x+t+1)=0.
$$

It is easy to see that the two roots of the equation are $x=t$ and $x=-t-1$, confirming that the equation indeed has two integer roots.
From the above two aspects, we have
$a=-\frac{2006}{t^{2}+t+1}$ (where $t$ is any integer).
Noting that the value of the function $y=t^{2}+t+1$ is a positive integer (since the variable $t$ is an integer), and the axis of symmetry is the line $t=-\frac{1}{2}$, we have
$$
a=-\frac{2006}{t^{2}+t+1}(t=0,1, \cdots).
$$

Furthermore, by the absolute difference of the two roots being no less than $2 k-1$, we get
$$
t-(-t-1) \geqslant 2 k-1,
$$

which simplifies to $t \geqslant k-1$.
Thus, $t=k-1, k, \cdots$.
In summary, the real number $a$ is
$$
a=-\frac{2006}{t^{2}+t+1}(t=k-1, k, \cdots).
$$