I understood what they are trying to say.

Suppose we wrote $9$ numbers $1,2,3,4,5,6,7,8,9$ each on different piece of paper and then put all those papers in a box and we shuffle those pieces.

Now how many minimum number of pieces we need to take out from that box to ensure that the number written on any $2$ of the pieces that we are taking out from the box results in $even$ number without looking at the numbers that we are picking out.

here the name of the box is $S$. i.e. box=set.

The answer will $3$.

**ROUND 1**

suppose we take any number from the box then it may be $even$ or $odd$.

if $even$ say $4$ then our job is done.

if $odd$ say $9$ then we need to pick another number from the box and so we go for Round 2.

**ROUND 2**

In the pick for 2nd time

if $odd$ say $7$ comes then then our job is done. since $9+7 =16 = even$

if $even$ number comes say $4$ then we need to take out another number which can again be $even$ or $odd$. ($\because$ $4+9=13$ is odd). i.e. we need to go for Round 3.

**ROUND 3**

in the 3rd pick

if an even number say $2$ comes then $2+4=6=even$

if an odd number say $5$ comes then $5+9=14=even$

We don’t need to go for round $4$

i.e. we need to take out atleast $3$ numbers from the box to ensure that any of those $2$ numbers result in $even$ sum.

hence $3$ is the answer.