No, check$\pmod2$.

Indeed, no matter what signs you use, the expression is equivalent to $1 \pmod{2}$, impossible. Another way to think about it is to originally assume that all the $\star$'s are $+$ signs. This gives a sum of $45$. For every number we change to a minus sign, we would need to decrease this sum by twice the original number. For example, if we made $8$ into $-8$, we would subtract $8(2)=16$ from the sum. But no matter how many numbers you choose to turn into minuses, we are subtracting even numbers only and can never achieve zero. If these numbers are $a_1$, $a_2$, ..., $a_n$, then we would have $45-\underbrace{2\left(a_1+a_2+ \dots + a_n\right)}_{\text{always even}} \equiv 1 \pmod{2}$.