For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case $N=1012$):
Let $N$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{2N-1}$ be positive integers such that
- $a_{1}, a_{2}, \ldots, a_{2N-1}$ is a permutation of $1,2, \ldots, 2N-1$, and
- $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2N-2}-a_{2N-1}\right|$ is a permutation of $1,2, \ldots, 2N-2$.
Then $a_{1}+a_{2N-1} \geqslant N+1$ and hence $\max \left(a_{1}, a_{2N-1}\right) \geqslant\left\lceil\frac{N+1}{2}\right\rceil$.

Now we proceed to the proof of the generalised statement. We introduce the notion of score of a number $a \in\{1,2, \ldots, 2N-1\}$. The score of $a$ is defined to be
$$s(a):=|a-N| .$$
Note that, by the triangle inequality,
$$|a-b| \leqslant|a-N|+|N-b|=s(a)+s(b)$$
Considering the sum $\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{2N-2}-a_{2N-1}\right|$, we find that
$$\begin{aligned} (N-1)(2N-1) & =\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{2N-2}-a_{2N-1}\right| \\ & \leqslant 2\left(s\left(a_{1}\right)+s\left(a_{2}\right)+\cdots+s\left(a_{2N-1}\right)\right)-\left(s\left(a_{1}\right)+s\left(a_{2N-1}\right)\right) \\ & =2N(N-1)-\left(s\left(a_{1}\right)+s\left(a_{2N-1}\right)\right) \end{aligned}$$
For the last equality we used that the numbers $s\left(a_{1}\right), s\left(a_{2}\right), \ldots, s\left(a_{2N-1}\right)$ are a permutation of $0,1,1,2,2, \ldots, N-1, N-1$.
Hence, $s\left(a_{1}\right)+s\left(a_{2N-1}\right) \leqslant 2N(N-1)-(N-1)(2N-1)=N-1$. We conclude that
$$\left(N-a_{1}\right)+\left(N-a_{2N-1}\right) \leqslant s\left(a_{1}\right)+s\left(a_{2N-1}\right) \leqslant N-1,$$
which implies $a_{1}+a_{2N-1} \geqslant N+1$.