1. $m\ge\left\lfloor\frac n 2\right\rfloor+1$
$\because\begin{cases}|r_1-r_n|+|c_1-c_n|\ge n-1\\|r_1-r_n|\le m-1\\|c_1-c_n|\le m-1\end{cases}$
$\therefore m-1+m-1\ge n-1$
$\therefore m\ge\frac{n+1}2$
$\because m\text{是整数}$
$\therefore m\ge\left\lfloor\frac n 2\right\rfloor+1$
2. $m$ 可以取到 $\left\lfloor\frac n 2\right\rfloor+1$
在每一斜行放一颗棋子即可，即：$r_i+c_i=i+1$。因为 $|r_i-r_j|+|c_i-c_j|\ge|r_i+c_i-r_j-c_j|$。
```cpp
#include <cstdio>
using namespace std;
int main()
{
int n, i, ans;
scanf("%d", &n);
ans = n / 2 + 1;
printf("%d", ans);
for (i = 1; i <= ans; ++i) printf("\n%d 1", i);
for (i = 2; i <= n - ans + 1; ++i) printf("\n%d %d", ans, i);
return 0;
}
```