Even or Odd?

Firstly, we need to know 2 of these crucial facts of the day! $n\equiv -n \pmod {2}$ and $\vert n \vert \equiv n \pmod {2}$ Now, $\begin{aligned} x&=|m_1-m_2|\\&\equiv |m_1+m_2|\\&\equiv m_1+m_2\\&=|l_1-l_2|+|l_3-l_4| \\&\equiv l_1+l_2+l_3+l_4\\&=|k_1-k_2|+|k_3-k_4|+|k_5-k_6|+|k_7-k_8| \\&\equiv k_1+k_2+k_3+k_4+k_5+k_6+k_7+k_8 \\&=\ldots \\&\equiv b_1+b_2+b_3+\ldots+b_{4095}+b_{4096} \\&=|a_1-a_2|+|a_3-a_4|+\ldots+|a_{8191}-a_{8192}| \\&\equiv a_1+a_2+a_3+\ldots+a_{8192} \\&=1+2+3+\ldots+8192 \\&\equiv 0\pmod {2} \end{aligned}$ Hence, $x$ must be even.

The initial sequence has as many even numbers as odd numbers (and the quantity of both is even). When an odd number meets an odd number, they will result in even number, so the amount of odd numbers will decrease by $2$ (or a multiple of $2$ , if this happens multiple times between two sequences). When an odd number meets an even number, they will result in an odd number, so the amount of odd numbers will stay the same. With these results, we proved that the amount of odd numbers will always be even.

This implies that the last number is even, because otherwise there would be an odd amount (in this case, $1$ ) of odd numbers. Also, there would need to be an odd amount of odd numbers in all of the previous sequences, among the which the initial sequence of the problem would obviously not be.