The largest such $k$ is $k=67$. Indeed, let us replace $100$ by $n$. By the condition, we have $x_{j+d} \ge x_j+d$ for all $j,d \ge 1$ and hence \[x_k+x_{k+1}+\dots+x_n \ge x_{k-1}+x_{k-2}+\dots+x_{2k-1-n}+(n+1-k)^2.\]Also note that this sharp since we can make the sequence $x_{2k-1-n},\dots,x_n$ just grow linearly with slope $1$. So the problem is equivalent to whether we have \[x_1+x_2+\dots+x_{2k-2-n} \le (n+1-k)^2\]for all such sequences. But the LHS can be at most \[2+4+\dots+2(2k-2-n)=(2k-2-n)(2k-1-n)\]and again this is clearly sharp. So the condition is satisfied iff $(2k-2-n)(2k-1-n) \le (n+1-k)^2$ which in turn is clearly satisfied iff $2k-1-n \le n+1-k$ and hence iff $k \le \frac{2n+2}{3}$. Inserting $n=100$ we get that the condition holds iff $k \le 67$.