Book has pages, right?

Sum of page numbers of 1 page is odd. So 1 page can be there, 2 cant, 3 can and so on since 101 is odd

For 1 page this isnt possible sum can be 51+52=103 or 49+50=99, not in between

For 3, we let the page numbers be $2a+1+2a+2+2b+1+2b+2+2c+1+2c+2=101$

So $a+b+c=23$

$a <b<c$ as they cant be equal and we have to not count duplicate solutions

So we have $a=1, b=2,3,...10$

$a=2, b=3,4,.....10$

$a=3, b= 4,5,...9$

$a=4, b=5,6,7,8,9$

$a=5, b=6,7,8$

$a=6, b=7,8$

c is fixed in every case so we have $9+8+6+5+3+2=33$ cases

For 5, we have $2a+1+2a+2+.....+2e+1+1+2e+2=101$ or $a+b+c+d+e=21.5$ not possible

For 7 we have $2a+1....+2g+2=101$ or $a+b+...+g=20$ but for 7 different integers minimum sum is 28. So further cases not possible with same reasoning