If x and y are consecutive odd integers such that x < y

The question states that x and y are consecutive odd integers such that x < y.
We are asked to find the value of 'x+y'.
Let's look at the given statements.

Statement 1 says that the product of x and y is negative.
If the odd integers x and y are both positive, their product will be positive. (E.g. 3 and 5, 19 and 21, etc)
Similarly, if x and y are both negative, their product will still be positive. (-1 and -3, -13 and -15, etc)
Hence, the only possible numbers which satisfy the given condition are 1 and -1 i.e. one negative and other positive number. Here x = -1 and y = 1 since x < y.
1 and -1 are consecutive odd integers such that their product is -1 which is negative.
Hence we can find the sum of these 2 odd integers as -1 +1 = 0.
Hence this statement by itself is sufficient.

Statement 2 says that the sum x + y is the square of an integer.
Let's consider a few examples which satisfy this condition.
Let x = 1 and y = 3. Their sum = 1+3 = 4 (which is the square of 2 or -2)
If x = 7 and y = 9, then their sum 7+9 = 16 is the square of 4 or -4.
We see that there can be a number of possible combinations of numbers which satisfy this condition.
Hence, this statement by itself is insufficient.

As statement 1 is sufficient and statement 2 is not, the correct answer is option A.

Hope this helps