Let the elements of $X$ be $a_1<a_2<...<a_8$ and denote the seven successive differences by $d_i=a_{i+1}-a_i.$
Consider the subsets of size $4$ which contain either $2$ or $3$ elements of $\{a_5,a_6,a_7,a_8\}$. There are $$\begin{pmatrix}4\\1\\\end{pmatrix}\begin{pmatrix}4\\3\\\end{pmatrix}+\begin{pmatrix}4\\2\\\end{pmatrix}\begin{pmatrix}4\\2\\\end{pmatrix}=52$$ of these subsets and the possible sums of their elements range from $a_1+a_2+a_5+a_6$ to $a_4+a_6+a_7+a_8$. So, by the pigeon-hole principle, we are finished unless $$a_4+a_6+a_7+a_8-(a_1+a_2+a_5+a_6)+1\ge 52$$ $$\text {i.e.} 2(a_8-a_1)\ge51+d_1+d_4+d_7.$$
Since $a_8-a_1\le 29$ we must have $d_1+d_4+d_7\le7$. Using the observations given below, $d_1,d_4,d_7$ are all different and no two can add to the third and so $\{d_1,d_4,d_7\}=\{1,2,4\}$ and $\{a_1,a_{8}\}=\{1,30\}.$
Some observations about the $d_i$.
(1) Any two non-adjacent differences are unequal.
(2) Given three non-adjacent differences, none is the sum of the other two.
(3) Given two adjacent differences, the sum of these differences can replace one of the differences in observations (1) and (2). (We still require the 'combined difference' to be non-adjacent to the other differences involved.)
The proofs of these are all elementary and of the same form. As an example, suppose we have $d_2+d_3=d_5+d_7$, which is a combination of (2) and (3). Then $$a_4-a_2=a_6-a_5+a_8-a_7.$$
The sets $\{a_4,a_5,a_7\}$ and $\{a_2,a_6,a_8\}$ then have the same sum and $a_1$, say, can be added to each.
To return to the main proof where we know that the differences $\{d_1,d_4,d_7\}=\{1,2,4\}$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $1$. Then, by the observations, $\{d,d+1\}\cap\{2,4,6\}$ is empty. So $d\ge7$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $2$. Then, by the observations, $\{d,d+2\}\cap\{1,3,4,5\}$ is empty. So $d\ge6$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $4$. Then, again by the observations, $\{d\}\cap\{1,2,3\}$ is empty. So $d\ge4$.
The sum of the differences (which is $29$) is now at least $(1+2+4)+(7+6+4)+d$, where $d$ is the 'other' difference adjacent to $d_4$. Therefore $d_4=4$ and the two differences adjacent to it (which cannot be equal) are $4$ and $5$. The differences adjacent to the differences of $1$ and $2$ are thus forced to be $7$ and $6$, respectively. Then $a_1+a_8=a_3+a_5$ and we are finished.
One way to see this combinatorially is to consider the $10$ elements in some order. With each of the first $9$ elements, you have a free choice of either including it or not including it, for $2$ choices each. However, there's no choice for the $10$th element, since to ensure the total # of elements in the set is odd, this element must be included if the total # so far is even, and not included if the total # is odd. Thus, since there are $2$ choices for each of the first $9$ elements, this means there are $2^9$ choices overall.
Best Answer
To make such a subset, for each of pair that sums to $n+1$, you have to decide between $3$ options: Include the only lower number, include only the higher number, or not include any of them. This gives a total of $$ 3^{\lfloor n/2\rfloor} $$ options.
When $n$ is odd, you also have the number $\frac{n+1}2$ which is not part of any pair. This one you can freely decide whether to include.
So all in all, we get $$ \text{Number of such subsets} = \cases{3^{n/2} & if $n$ is even\\ 2\cdot 3^{(n-1)/2} & if $n$ is odd} $$