20. For every real $x$ we shall denote by $\lfloor x\rfloor$ and $\lceil x\rceil$ the greatest integer less than or equal to $x$ and the smallest integer greater than or equal to $x$ respectively. The condition $c_{i}+n k_{i} \in[1-n, n]$ is equivalent to $k_{i} \in I_{i}=$ $\left[\frac{1-c_{i}}{n}-1,1-\frac{c_{i}}{n}\right]$. For every $c_{i}$, this interval contains two integers (not necessarily distinct), namely $p_{i}=\left\lceil\frac{1-c_{i}}{n}-1\right\rceil \leq q_{i}=\left\lfloor 1-\frac{c_{i}}{n}\right\rfloor$. In order to show that there exist integers $k_{i} \in I_{i}$ with $\sum_{i=1}^{n} k_{i}=0$, it is sufficient to show that $\sum_{i=1}^{n} p_{i} \leq 0 \leq \sum_{i=1}^{n} q_{i}$. Since $p_{i}-\frac{c_{i}}{n}$ implies
$$\sum_{i=1}^{n} q_{i}>-\sum_{i=1}^{n} \frac{c_{i}}{n} \geq-1$$
which leads to $\sum_{i=1}^{n} q_{i} \geq 0$. The proof is complete.