Lucky Seven

Let $A_{i}$ be a subset of X such that $A_{i}$ = $7n+i$ $n \in \mathbb{I}$

$A_{0}=\{7,14,21,....98\}$

$A_{1}=\{1,8,15,.....99\}$

$A_{2}=\{2,9,16,....100\}$

$A_{3}=\{3,10,17,....94\}$

$A_{4}=\{4,11,18,....95\}$

$A_{5}=\{5,12,19,....96\}$

$A_{6}=\{6,13,20,....97\}$

Number of elements in $A_{0}=14$

$A_{1}=15$

$A_{2}=15$

$A_{3}=14$

$A_{4}=14$

$A_{5}=14$

$A_{6}=14$

If we select an element from $A_{1}$ , no element from $A_{6}$ can be selected as the sum of the two elements would be divisible by 7.

So for maximum possible elements Y should have : 1 element from $A_{0}$ ,15 elements from $A_{1}$ , 15 elements from $A_{2}$ and 14 elements from either $A_{3}$ or $A_{4}$ .

So the maximum possible number of elements $=1+15+15+14 =\boxed{45}$