Obviously all of these $22$ integers are elements of the set $\{1, -1\}$, and clearly there are an even number of $-1$s and an even number of $1$s. For the sum to equal $0$, there must be an equal number of $-1$s and $1$s, so the number of $1$s/$-1$s is $2k$ for some $k$. This means there are total of $4k$ integers, this contradicts the fact that $v_2(22) =1$. So it cannot be $0$.

all the integers are equal to 1 or -1.if there sum is 0 , then there are 11 1's and 11 (-1)'s .then clearly we have a contradiction.

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