Fix $k > 0$ for $\lceil \frac{2k+1}{3} \rceil \le j \le k$,

$$ \sum_{i = 0}^{2j-k-1} \binom{j}{i} + \sum_{b = \lceil \frac{k+1}{2} \rceil}^{\lfloor \frac{2k-j}{2} \rfloor} \left( \sum_{l = 0}^{2b-k-1} \binom{b}{l} \right) \binom{j-b-1}{2j-(2k+1)+b}$$

I think that this expression (mod 2) should be 1 when $j=k$ and 0 otherwise. I can show that it is 1 when $j = k$ and have checked the other cases for relatively small values of $k$, but have been unable to show in general that when $j \not= k$ the expression is 0 mod 2.