One can rather easily show that $E\left[\sum\limits_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})\right] = -T + W_T^2$.
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 $E\left[\sum\limits_{i = 0}^{i = n - 1}W_{t_i}(W_{t_{i + 1}} - W_{t_i})\right] = \sum\limits_{i = 0}^{i = n - 1}E\left[W_{t_{i}}\right]E\left[W_{t_{i + 1}} - W_{t_i}\right] = 0$?
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.
You have an error: you find that the expectation is $-T+E[W_T^2]=-T+T=0$. That is, your observation is correct. We usually view this as arising from the fact that in the Ito convention, $\int_0^t W_s dW_s$ is a martingale. (So are the integrals of "most" functions against $W_t$.)