To solve the problem, we need to evaluate the summation of the greatest integer function of logarithm base 2 from \( r = 1 \) to \( n \) and set it equal to 2010.
### Step-by-Step Solution:
1. **Understanding the Summation**:
We need to evaluate:
\[
\sum_{r=1}^{n} \lfloor \log_2 r \rfloor = 2010
\]
The function \( \lfloor \log_2 r \rfloor \) gives the greatest integer less than or equal to \( \log_2 r \). This function changes its value at powers of 2.
2. **Identifying the Values of \( \lfloor \log_2 r \rfloor \)**:
- For \( r = 1 \): \( \lfloor \log_2 1 \rfloor = 0 \)
- For \( r = 2, 3 \): \( \lfloor \log_2 r \rfloor = 1 \)
- For \( r = 4, 5, 6, 7 \): \( \lfloor \log_2 r \rfloor = 2 \)
- For \( r = 8, 9, 10, 11, 12, 13, 14, 15 \): \( \lfloor \log_2 r \rfloor = 3 \)
- For \( r = 16, 17, \ldots, 31 \): \( \lfloor \log_2 r \rfloor = 4 \)
- And so on...
3. **Counting Contributions**:
The contributions to the sum can be counted based on the ranges defined by powers of 2:
- From \( 1 \) to \( 1 \): contributes \( 0 \times 1 = 0 \)
- From \( 2 \) to \( 3 \): contributes \( 1 \times 2 = 2 \)
- From \( 4 \) to \( 7 \): contributes \( 2 \times 4 = 8 \)
- From \( 8 \) to \( 15 \): contributes \( 3 \times 8 = 24 \)
- From \( 16 \) to \( 31 \): contributes \( 4 \times 16 = 64 \)
- From \( 32 \) to \( 63 \): contributes \( 5 \times 32 = 160 \)
- From \( 64 \) to \( 127 \): contributes \( 6 \times 64 = 384 \)
- From \( 128 \) to \( 255 \): contributes \( 7 \times 128 = 896 \)
- From \( 256 \) to \( 511 \): contributes \( 8 \times 256 = 2048 \) (but we will not reach this far since we need to total 2010)
4. **Summing Contributions**:
Now we sum these contributions until we reach or exceed 2010:
\[
0 + 2 + 8 + 24 + 64 + 160 + 384 + 896 = 1534
\]
This means we have accounted for \( 1534 \) up to \( r = 255 \).
5. **Finding Remaining Value**:
We need to reach \( 2010 \):
\[
2010 - 1534 = 476
\]
Now, for \( r = 256 \) and onwards, \( \lfloor \log_2 r \rfloor = 8 \).
6. **Calculating Additional Terms**:
We can add \( 8 \) for each additional \( r \):
\[
8k = 476 \implies k = \frac{476}{8} = 59.5
\]
Since \( k \) must be an integer, we can only take \( k = 59 \) which gives:
\[
8 \times 59 = 472
\]
Thus, we can take \( r \) from \( 256 \) to \( 314 \) (59 terms).
7. **Finding \( n \)**:
Therefore, \( n = 255 + 59 = 314 \).
8. **Sum of Digits of \( n \)**:
Now, we find the sum of the digits of \( n = 314 \):
\[
3 + 1 + 4 = 8
\]
### Final Answer:
The sum of the digits of \( n \) is \( \boxed{8} \).
