Inequality about unit vectors $2n$ unit vectors with non-negative $y$-coordinates are given. Prove that if the sum of $x$-coordinates is an odd integer, then the sum of $y$-coordinates is at least $1$.
$n=1$ case is easy; I also succeeded to prove $n=2$ case, but I cannot generalize this to every cases.


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*Source: India Postal Coaching 2014.

 A: Let the vectors be $u_k = (x_k,y_k)$ for $1\leqslant k \leqslant 2n$.
Consider first the case where $x_k \geqslant 0$ for all $k$. Using $\lvert y_k\rvert \geqslant 1 - \lvert x_k\rvert$ for unit vectors, we have
$$\sum_{k = 1}^{2n} y_k \geqslant \sum_{k = 1}^{2n} (1 - x_k) = 2n - \sum_{k = 1}^{2n} x_k,$$
so $\sum\limits_{k = 1}^{2n} y_k$ is a non-negative odd integer, hence $\geqslant 1$. The case of $x_k \leqslant 0$ for all $k$ is dealt with by symmetry.
If there is a pair $(k,m)$ with $-1 < x_k < 0 < x_m < 1$, we can replace the corresponding pair of unit vectors with either $u_k' = (-1,0), \, u_m' = (x_k + x_m + 1, y_k')$ or $u_k' = (x_k + x_m -1, y_m'),\, u_m' = (1,0)$, depending on whether $\lvert x_k\rvert \geqslant x_m$ or $\lvert x_k\rvert < x_m$. That doesn't change the sum of the $x$-coordinates, and makes the sum of $y$-coordinates smaller.
We iterate that replacement while there are such pairs, and arrive at a situation where there are no $k$ with $0 < x_k < 1$, or no $k$ with $-1 < x_k < 0$.
Now replace either all vectors $u_k = (1,0)$ with $(-1,0)$, or all vectors $u_k = (-1,0)$ with $(1,0)$. That changes the sum of $x$-coordinates by an even integer, and doesn't change the sum of $y$-coordinates and leads us to the first case where either $x_k \geqslant 0$ for all $k$ or $x_k \leqslant 0$ for all $k$. It follows that the sum of the $y$-coordinates of the modified vectors is $\geqslant 1$, but since all modifications did not increase any $y$-coordinate, the sum of original $y$-coordinates was $\geqslant 1$.
