Here's my solution:

Each sheet has two page numbers (one front and one back). So there are 400 sheets. If we think of it as a three digit number (starting with 001), then the first digit on the remaining pages will be either a 0, 2, or from 4 to 7 (800, the last page number, is on the backside of a sheet), thats six different possible digits. The second digit will be 0, 2, or from 4 to 9, eight different possible digits. and the third digit will be 5&6, 7&8, or 9&0, 3 different possible digits (counting them in pairs). This makes 6 times 8 times 3, or 144 sheets.

But this still includes the sheets where a 1 or 3 appears on the backside but not the front (like sheet 29-30 or 99-100). This happens in one of two ways, either with the first or the second digit (the third digit will never be odd on the backside of the sheet).

For the first digit, its simply 99-100 and 299-300 that still need removed. For the second digit, its the 09-10 and 29-30 for every possible first digit. That is, 6 possible first digits times 2 possible second digits (the third is always a 9 in this case). So 6 times 2 is 12 plus the 2 from the first digit is 14 sheets that still need removed.

144 - 14 = 130 sheets remaining.

Is case you got lost above, here's a breakdown of the remaining sheets:

1st | 2nd | 3rd

digit | digit | digit

is | is | is

-------------------------------

0,2,4-7 | 0,2,4-9 | 5&6, 7&8 = 6 * 8 * 2 = 96

4-7 | 4-9 | 9&0 = 4 * 6 * 1 = 24

0,2 | 4-8 | 9&0 = 2 * 5 * 1 = 10

96 + 24 + 10 = 130

So thats 130 pages remaining (pages meaning sheets), or 260 by page numbers.