One can rather easily show that E[$\sum_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})]$ = -T + $W_T^2$.
What I'm confused about is why we can't simply say that for each i, $W_{t_{i}}$ is independent of $(W_{t_{i + 1}} - W_{t_i})$, so that upon interchanging sums and expectations, and using independence we have E[$\sum_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})]$ = $\sum_{i = 0}^{i = n - 1}E[W_{t_{i}}]E[W_{t_{i + 1}} - W_{t_i}]$ = 0?
In this problem, we have naturally partitioned an interval as $0 = t_{0} < t_{1} < ... < t_{n} = T$, and $W_{t}$ is a Brownian motion.