A subset of $ A$ with the desired property cannot have 8 elements. In a $ 8$ element set,since $ a_6$ can be filled in 7 ways, by pigeon principle we can find two distinct elements $ (a_1,a_2,\ldots,a_8)$ and $ (b_1,b_2,\ldots,b_n)$ we can have only $ a_7\neq b_7$ and $ a_8\neq b_8$. A construct of $ X$ with 7 elements: $ 1,1,1,1,1,1,1,1$ $ 1,1,1,1,1,2,2,2$ . . . . $ 1,1,1,1,1,7,7,7$

Maybe I misundustood the problem, but if you add: $ 1,2,1,1,1,7,8,9$ you have eight members in $ X$

Yes, we can add the 8 th element in this case. Then there is something wrong with what I wrote.

I wrote: ... we can have only $ a_7\neq b_7$ and $ a_8\neq b_8$. I see here is where I slipped. First 6 elements can be filled in $ 2.3.4.5.6.7=5040$ ways. In a set with $ 5041$ elements,by pigeon hole principle we can find $ (a_1,\ldots,a_8)$ and $ (b_1,\ldots,b_8)$ so that $ a_i=b_i\quad \forall 1\leq i\leq 6$ qed ?

Clearly you prove that the number is lower than 5041, however is it possible to find a 5040-example?

Actually,no clear proof was given here,so here is my proof,i don't know official,but maybe they are the same: Answer:$ 7!$ Solution: In fact,them most hard part of the problem is to construct correct example,when the right answer achieved and to prove that it is correct indeed. But at first we will proof that $ |X|\leq 7!$. Let $ X$ be sparse subset of $ A$ and $ \overline{x}$ be a sequence of natural numbers $ (a_1,a_2,\dots,a_n)$,if i write $ \overline{x}\in X$,that means $ \overline{x} = (a_1,a_2,\dots,a_8)$. Consider $ X' = \{(a_1,a_2,\dots,a_6)$: such that for some $ a_7,a_8\in\mathbb{N}$,element $ (a_1,a_2,\dots,a_8)\in X)\}$ Let $ \overline{x},\overline{y}\in X'$,then $ \overline{x}\neq\overline{y}$,otherwise there will be less than $ 3$ indices,in $ x = \overline{x}(x_7,x_8),y = \overline{y}(y_7,y_8)\in X$,such that $ x_i\neq y_i$,contradiction.Hence $ \overline{x}\neq\overline{y}$. So $ |X| = |X'|\leq 2\cdot 3\cdot 4\dots\cdot 7 = 7!$. Let $ A' = \{(a_1,a_2\dots,a_6)$,$ 1\leq a_i\leq i + 1,a_i\in\mathbb{Z}\}$.Thus $ |A'| = 7!$. Now for each $ \overline{a'}\in A'$,we construct $ \overline{a} = \overline{a'}(a_7,a_8)$,such that $ a_7 = a_1 + a_2 + \dots + a_6 (mod 7)$,and $ a_8 = \sum_{i = 1}^{6} i\cdot a_i (mod 7)$. The last step is to prove that $ X = \{\overline{a}: \overline{a'}\in A'\}$-is a sparse subset of $ A$. Consider two elements of $ X$,$ \overline{a},\overline{b}$. It is easy to prove that there are at least three $ 3$ indices,such that $ a_i\neq b_i$,just consider three cases, 1. $ \overline{a'},\overline{b'}$ have three diffrent indices. 2. $ \overline{a'},\overline{b'}$ have two diffrent indices. 3. $ \overline{a'},\overline{b'}$ have one diffrent indice. So the problem is proved.