[paulfr]

There are no absolute values in the summands, so in your example pairs of consecutive terms sum to 0 when you choose b = 2.

The theorem seems entirely correct to me. You can prove it with these sub-steps:

(1) the set of all j + a_j is the set of nonnegative integers minus a finite number of gaps

(2) thus for large enough n you can express \sum_{j=1}^n (j + a_j) as a quadratic function of n, plus a residual term e(n) of magnitude at most 1007^2/2

(3) more precisely, \sum_{j=m+1}^n a_j = g (n-m) + e(n) - e(m) where g is the number of gaps in (1)

Then choosing b = g solves the problem.

Hope that helps.