2022 CGMO

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, $$\begin{array}{rl}\sum _{j=67}^{100}{x}_j-\sum _{i=0}^{66}{x}_i& =\sum _{j=1}^{34}(x_{66+j}-x_{32+j})-\sum _{i=1}^{32}{x}_i\\ & \ge 34^2-\sum _{i=1}^{32}2i\\ & =34^2-32\times 33>0.\end{array}$$ In conclusion, the largest $k$ we seek is $67$. □