Starting with the "1" in the centre, the spiral of consecutive integers continues, as shown. What is the sum of the number that appears directly above 2007 and the number that appears directly below 2007 ?
(A) 4014
(B) 4016
(C) 4018
(D) 4020
(E) 4022

| 17 | 16 | 15 | 14 | 13 |
| :---: | :---: | :---: | :---: | :---: |
| $\downarrow$ | 5 | 4 | 3 | 12 |
|  | 6 | 1 | 2 | 11 |
|  | 7 | 8 | 9 | 10 |
|  |  |  |  |  |
First, we complete the next several spaces in the spiral to try to get a better sense of the pattern:

| 17 | 16 | 15 | 14 | 13 |  |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 18 | 5 | 4 | 3 | 12 |  |
| 19 | 6 | 1 | 2 | 11 |  |
| 20 | 7 | 8 | 9 | 10 |  |
| 21 | 22 | 23 | 24 | 25 | 26 |

We notice from this extended spiral that the odd perfect squares lie on a diagonal extending down and to the right from the 1, since 1,9,25 and so on will complete a square of numbers when they are written. (Try blocking out the numbers larger than each of these to see this.) This pattern does continue since when each of these odd perfect squares is reached, the number of spaces up to that point in the sequence actually does form a square.

The first odd perfect square larger than 2007 is $45^{2}=2025$.

2025 will lie 18 spaces to the left of 2007 in this row. (The row with 2025 will actually be long enough to be able to move 18 spaces to the left from 2025.)

The odd perfect square before 2025 is $43^{2}=1849$, so 1850 will be the number directly above 2025, as the row containing 1849 will continue one more space to the right before turning up.

Since 1850 is directly above 2025 , then 1832 is directly above 2007 .

The odd perfect square after 2025 is $47^{2}=2209$, so 2208 will be the number directly below 2025 , since 2209 will be one space to the right and one down.

Since 2208 is directly below 2025 , then 2190 is directly below 2007 .

Therefore, the sum of the numbers directly above and below 2007 is $1832+2190=4022$.

ANSWER: (E)