4. (5 points) The last three digits of the expression $1 \times 1+11 \times 11+111 \times 111+\cdots+111 \cdots 111$ (2010 1’s) $\times 111 \cdots 111$ (2010 1’s) are $\qquad$ .
【Analysis】This problem seems difficult, but we can start by calculating the value of each multiplication expression from the first one in the formula to find a pattern: $1 \times 1=1, 11 \times 11=121, 111 \times 111=12321, 1111 \times 1111=1234321, 11111 \times$ $11111=123454321 \cdots$, their products are: $1, 121, 12321, 1234321, 123454321, 12345654321, \cdots$, From this, we can observe that, apart from the first two multiplication expressions whose products are 1 and 121, the last three digits of the products of the subsequent multiplication expressions are all 321. Based on this pattern, we can determine the last three digits of the product of this expression.

【Solution】Solution: By calculation, we get the products of each multiplication expression as follows: 1, 121, 12321, 1234321, 123454321, $12345654321, \cdots$,

From this, we can observe that, apart from the first two multiplication expressions whose products are 1 and 121, the last three digits of the products of the subsequent multiplication expressions are all 321;

Thus, the sum of the last three digits of each expression in the formula is:
$$
\begin{array}{l}
1 + 121 + 321 \times (2010 - 2) \\
= 122 + 64568 \\
= 644690 .
\end{array}
$$

Therefore, the last three digits of the result of the expression $1 \times 1 + 11 \times 11 + 111 \times 111 + \cdots + 111 \cdots 111$ (2010 ones) $\times 111 \cdots 111$ (2010 ones) are 690.

The answer is: 690.