Question 1046168
First of all, you must realize that, for ANY sheet of the notebook,
the front sheet will always be numbered ODD (and so the back page will always 
be numbered even.)

Thus, the number pages of the sheets will form a sequence that looks like this:

{{{O[1]}}}, {{{O[1]+1}}}, {{{O[2]}}}, {{{O[2]+1}}}, {{{O[3]}}}, {{{O[3]+1}}},..., {{{O[42]}}}, {{{O[42]+1}}}, {{{O[43]}}}, {{{O[43]+1}}},

where {{{O[k]}}} denotes the odd number paging of the front page of the kth sheet.

Suppose we sum all of these pagings.  The sum would be 

{{{sum(2*O[k], k=1,43)+43}}}.

Letting this equal 2016, we get

{{{sum(2*O[k], k=1,43)+43 = 2016}}}.

===> {{{2sum(O[k], k=1,43)= 1973}}}.

Contradiction.

Therefore the sum CANNOT be 2016.