Proof of necessity: Assume that there exists $\{x_n\}$ satisfying (1)–(3). Notice that the expression in (3) can be written as
$$x_n - x_{n-1} = \sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}), \quad n \in \mathbb{N}.$$
As $x_0 = 0$, we then have
$$\begin{align*} x_n &= \sum_{l=1}^{n} (x_l - x_{l-1}) = \sum_{l=1}^{n} \sum_{k=1}^{2008} a_k (x_{l+k} - x_{l+k-1}) \\ &= \sum_{k=1}^{2008} \sum_{l=1}^{n} a_k (x_{l+k} - x_{l+k-1}) \\ &= \sum_{k=1}^{2008} a_k (x_{n+k} - x_k). \end{align*}$$
From (2) we are able to define $b = \lim_{n \to \infty} x_n$. Let $n \to \infty$ in the above expression. Then we have
$$b = \sum_{k=1}^{2008} a_k (b - x_k) = b \sum_{k=1}^{2008} a_k - \sum_{k=1}^{2008} a_k x_k \\ < b \sum_{k=1}^{2008} a_k.$$
Therefore, $\sum_{k=1}^{2008} a_k > 1$.

Proof of sufficiency: Assume that $\sum_{k=1}^{2008} a_k > 1$. Define a polynomial function by
$$f(s) = -1 + \sum_{k=1}^{2008} a_k s^k, \quad s \in [0, 1].$$
$f(s)$ is strictly increasing on the interval $[0, 1]$. In the meantime,
$$f(0) = -1 < 0, \quad f(1) = -1 + \sum_{k=1}^{2008} a_k > 0,$$
and there then exists a unique $0 < s_0 < 1$ such that $f(s_0) = 0$.
Now, define $x_n = \sum_{k=1}^{n} s_0^k$, $n \in \mathbb{N}$. It is easy to see that $\{x_n\}$ satisfies (1) and
$$x_n = \sum_{k=1}^{n} s_0^k = \frac{s_0 - s_0^{n+1}}{1 - s_0}.$$
On the other hand, $\lim_{n \to \infty} s_0^{n+1} = 0$ as $0 < s_0 < 1$. Then we have
$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{s_0 - s_0^{n+1}}{1 - s_0} = \frac{s_0}{1 - s_0}.$$
This means that $\{x_n\}$ satisfies (2). Finally, we have
$$0 = f(s_0) = -1 + \sum_{k=1}^{2008} a_k s_0^k.$$
That is to say, $\sum_{k=1}^{2008} a_k s_0^k = 1$. Then we have
$$\begin{align*} x_n - x_{n-1} &= s_0^n \\ &= \left( \sum_{k=1}^{2008} a_k s_0^k \right) s_0^n \\ &= \sum_{k=1}^{2008} a_k s_0^{n+k} \\ &= \sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}). \end{align*}$$
Therefore, $\{x_n\}$ also satisfies (3). This completes the proof.