4. Arrange the squares of positive integers $1,2, \cdots$ in a sequence: $149162536496481100121144 \cdots$, the digit at the 1st position is 1, the digit at the 5th position is 6, the digit at the 10th position is 4, the digit at the 2008th position is $\qquad$.
4.1.
$1^{2}$ to $3^{2}$, each result occupies 1 digit, totaling $1 \times 3$ $=3$ digits;
$4^{2}$ to $9^{2}$, each result occupies 2 digits, totaling $2 \times 6$ $=12$ digits;
$10^{2}$ to $31^{2}$, each result occupies 3 digits, totaling $3 \times$ $22=66$ digits;
$32^{2}$ to $99^{2}$, each result occupies 4 digits, totaling $4 \times$ $68=272$ digits;
$100^{2}$ to $316^{2}$, each result occupies 5 digits, totaling $5 \times 217=1085$ digits;

At this point, we still need
$$
2008-(3+12+66+272+1085)=570
$$

digits.
$317^{2}$ to $411^{2}$, each result occupies 6 digits, totaling $6 \times 95=570$ digits.

Therefore, the digit at the 2008th position is precisely the unit digit of $411^{2}$, which is 1.