Assuming that the numbers are positive integers,
This is closely related to the Frobenius Coin Problem which says that there is a maximum number $\displaystyle F$ (called the Frobenius number) which is not representable. It is NP-Hard to find out the Frobenius number when there are at least $\displaystyle 3$ numbers.
For a formula like approach to determine if such a representation is possible or not, you can use generating functions, which can be used to give a pseudo polynomial time algorithm, polynomial in size $\displaystyle W = n_1 + n_2 + \cdots + n_k$.
If the numbers are $\displaystyle n_1, n_2, \dots, n_k$ and you need to see if they can be summed to $\displaystyle S$ then the number of ways it can be done is the coefficient of $\displaystyle x^S$ in
$$\displaystyle (1+x^{n_1} + x^{2n_1} + x^{3n_1} + \cdots )(1+ x^{n_2} + x^{2n_2} + x^{3n_2} + \cdots ) \cdots (1 + x^{n_k} + x^{2n_k} + x^{3n_k} + \cdots )$$
$$\displaystyle = \dfrac{1}{(1-x^{n_1})(1-x^{n_2}) \cdots (1-x^{n_k})}$$
Using partial fractions this can be written as
$$\displaystyle \sum_{j=1}^{m} \dfrac{C_j}{c_j - x}$$
where $\displaystyle C_j$ and $\displaystyle c_j$ are appropriate complex numbers and $\displaystyle m \le n_1 + n_2 + \cdots + n_k$.
The coefficient of $\displaystyle x^S$ is thus given by
$$\displaystyle \sum_{j=1}^{m} \dfrac{C_j}{c_j^{S+1}}$$
which you need to check is zero or not.
Of course, this might require quite precise floating point operations and does not actually tell you what numbers to choose.
OK, so here is a solution to (i) in case we know that when a page is torn out and odd and the following even page number is removed.
Given the book has $n$ pages the sum of the page numbers will be
$$
T_n=1+2+...+n=\frac{n(n+1)}{2}
$$
known as the $n$'th Triangular Number. So assume that page numbers $x$ and $x+1$ have been torn out where $x$ is odd. Then we can write
$$
\frac{105}{4}=\tfrac{1}{n-2}\left(T_n-(2x+1)\right)
$$
Plugging in the formula for $T_n$ from above and solving for $x$ then yields
$$
x=\frac{1}{4}n^2-\frac{103}{8}n+\frac{103}{4}
$$
which is a quadratic expression in $n$ with zeros $n=\frac{103\pm\sqrt{8961}}{4}$ which approximately is $n=2.08$ and $n=49.42$. On the other hand it is obvious that $x<n$ which then in turn yields a quadratic inequality in $n$ that can be solved to see that $n<\frac{111+\sqrt{10673}}{4}\approx 53.58$.
Having a slightly closer look at the expression for $x$ one realizes that $n-2$ must be divisible by $8$ for $x$ to be an integer. So unless we take $n=2$ (which actually works) we must have $49.42<n=50<53.58$ for $n$ to be an integer satisfying all requirements in that interval. Plugging $n=50$ into the expression for $x$ then yields $x=7,x+1=8$, and you can check as I did in the comments that the average of the remaining pages is really $\frac{105}{4}$. With $n=50$ one gets
$$
\sum_{i=1}^{10}\left\lfloor\frac{50+r}{r+1}\right\rfloor=105
$$
So option (C) answers the first question correctly.
Part (ii)
Still using $n=50$ we get $\lfloor\frac{50}{3}\rfloor=16$ so that the line is $x+y=16$ or $y=16-x$. Together with the axes $y=0$ and $x=0$ this encloses a triangle with $17$ lattice points on the line $y=16-x$ since you can start from $(0,16)$ and move down right step-by-step to 'visit' 17 lattice points before you hit the x-axis. Similarly you hit $16$ lattice points following the same procedure from $(0,15)$. With this we get the answer to (ii) which is:
$$
T_{17}=17+16+...+1=\frac{17\cdot 18}{2}=153
$$
So option (B) answers the second question correctly.
Best Answer
We have that $1+\cdots+100=5050.$ So, if the remaining pages sum $4949$ the pages which are not in the book add up $101.$ Moreover, if $2a-1$ is not in the book then $2a$ is also not in the book. They add up $4a-1.$ It's not possible to torn out only a page because $4a-1=101$ has no integer solution. If we turn out two pages we have to solve $4a-1+4b-1=101$ which also doesn't have integer solution. So, assume $n$ pages are torn out. We have $$4a_1-1+\cdots+4a_n-1=101 (\iff 4(a_1+\cdots+a_n)=101+n).$$ Thus the number of pages can be $3,7,11, \dots$ because $101+n$ must be a multiple of $4.$ Now, note that if $n\ge 7$ then $$\dfrac{n(n+1)}{2}=1+\cdots+n\le a_1+\cdots+a_n=\dfrac{101+n}{4}< \dfrac{n(n+1)}{2}$$ gives a contradiction. So, $n=3.$
Edit
Let's prove of the last inequality for $n\ge 7:$ $$\dfrac{101+n}{4}< \dfrac{n(n+1)}{2}\iff 2n^2+2n>101+n\iff 2n^2+n>101.$$ Now, $$n\ge 7\implies 2n^2+n\ge 98+7=105>101,$$ and we are done.
Edit 2
In order to determine the pages we have to solve $$a_1+a_2+a_3=26.$$ Then, the pages are $2a_1-1,2a_1;$ $2a_2-1,2a_2;$ and $2a_3-1,2a_3.$
Now, solving $a_1+a_2+a_3=26$ is to get the partitions of $26$ into distinct parts (that is, $a_1,a_2$ and $a_3$ are different). According to Wolfram Alpha there are $165$ solutions. (See https://www.wolframalpha.com/input/?i=PartitionsQ(26).)