The answer is actually $3.$ You can note that $b_i\neq b_{i+1}$ if $b_i=a_i+a_{i+1},$ thus for odd we know its at least 3 and for even at least 2. For even, note that if there are only two, so $b_1, b_2, b_1, b_2, dots$ the sum of all the $a_i$ is $\frac{nb_1}{2}=\frac{nb_2}{2},$ so contradiction. Therefore, $3$ is lower bound. You can make a construction using $0, 1, -1, 2, -2, \dots$