Solution:

Let $n = \overline{a_k a_{k-1} \ldots a_0}$ be an integer equal to the sum of squares of its own digits. Then
$$n = 10^k a_k + 10^{k-1} a_{k-1} + \cdots + a_0.$$
On the other hand,
$$a_0^2 + a_1^2 + \cdots + a_k^2 \leq 9^2 \cdot (k+1) = 81 \cdot (k+1) < 10^k \quad \text{for } k \geq 4.$$
(The last inequality is easy to prove by induction because it holds for $k=4$, and if it holds for some $k$, then
$$81 \cdot (k+2) = 81 \cdot (k+1) + 81 \leq 10^k + 81 < 10^k + 10^k < 10^{k+1}.$$
)

Therefore, for $k \geq 4$, the sum of squares of the digits is less than $10^k$, so $n$ cannot have more than $4$ digits. Thus, $n \leq 9999$.

Now, for $n \geq 2009$, $n$ must be a $4$-digit number. The maximum sum of squares of $4$ digits is $4 \times 81 = 324$, which is much less than $2009$. Therefore, no integer greater than $2008$ can be equal to the sum of squares of its digits.