, 2022

The answer is $67$.

On one hand, if $x_i = 2i$, $1 \le i \le 34$, $x_{34+j} = x_{34} + j = 68 + j$, $1 \le j \le 66$, then the sequence satisfies the conditions of the problem. We have
$$\sum_{j=68}^{100} x_j - \sum_{i=0}^{67} x_i = \sum_{j=1}^{33} (x_{67+j} - x_{34+j}) - \sum_{i=1}^{34} x_i = 33^2 - 34 \times 35 < 0.$$
This example shows that when $k \ge 68$, it does not satisfy the requirements.

On the other hand, for any sequence $x_1, x_2, \dots, x_{100}$ that satisfies the conditions of the problem, it is easy to know that $x_i \le 2i$, $1 \le i \le 100$. For $0 \le s < t \le 100$, there is the inequality $x_t - x_s \ge t - s$. Therefore,
$$\begin{aligned} \sum_{j=67}^{100} x_j - \sum_{i=0}^{66} x_i &= \sum_{j=1}^{34} (x_{66+j} - x_{32+j}) - \sum_{i=1}^{32} x_i \\ &\ge 34^2 - \sum_{i=1}^{32} 2i \\ &= 34^2 - 32 \times 33 > 0. \end{aligned}$$
In conclusion, the largest $k$ we seek is $67$. □