必要条件
一方面
$\sum_{i=1}^{N}(a_i + b_i) \le \sum_{i=1}^{N} c_i \implies 2\sum_{i=1}^{N} c_i \ge \sum_{i=1}^{N}(a_i + b_i + c_i) = \sum_{i=K}^{K+3N-1} i = \frac{3N(2K+3N-1)}{2}$
另一方面
$\sum_{i=1}^{N} c_i \le \sum_{i=K+2N}^{K+3N-1} i = \frac{N(2K+5N-1)}{2}$
$N(2K+5N-1) \le \frac{3N(2K+3N-1)}{2} \implies 2K - 1\le N$
此必要条件也可用另一种方法推导出来：
由于 $\sum_{i = 1}^{N} (a_i + b_i) \ge \sum_{i=K}^{K+2N-1} i $ 且 $\sum_{i=1}^{N} c_i \le \sum_{i = K+2N}^{K+3N-1} i$，因此 $\sum_{i = 1}^{N} (a_i + b_i) \le \sum_{i=1}^{N} c_i \implies \sum_{i=K}^{K+2N-1} i \le \sum_{i = K+2N}^{K+3N-1} i \implies 2K - 1\le N$。
构造
the pattern is $(x, y)$, $(x+2, y -1)$, ...
例子
$K = 2, N = 6$
\begin{matrix}
3 & 5 & 7 & 2 & 4 & 6 \\
10 & 9 & 8 & 13 &12 & 11 \\
\hline
14 & 15 & 16 & 17 & 18 & 19
\end{matrix}
$K = 2, N = 7$
\begin{matrix}
2 & 4 & 6 & 8 & 3 & 5 & 7 \\
15 & 14 & 13 & 12 & 11 & 10 & 9 \\
\hline
19 & 20 & 21 & 22 & 16 & 17 & 18
\end{matrix}