$C_1,C_2$ are positive constants, and $\alpha\in(0,1)$ is a constant too.
If for any $\epsilon>0$, we have $$ \sum_{i<j}(k_i-k_j)^2 \le C_1 \epsilon^{2\alpha} \\ \sum k_i \ge C_2 \epsilon^\frac{\alpha}{2} $$ How to show $k_i>0$ ?
This is my guess, when $\epsilon$ is small enough, the first inequality means the $k_i$s have little difference. Although the RHS of second inequality approached zero also, but it is of low order respect to $\epsilon^{2\alpha}$.
This question is origin from
Lin, Longzhi; Sesum, Natasa, Blow-up of the mean curvature at the first singular time of the mean curvature flow, Calc. Var. Partial Differ. Equ. 55, No. 3, Paper No. 65, 16 p. (2016). ZBL1343.53064.
OK, after your edit this makes more sense - if we normalize by setting $a_i = \epsilon^{-\alpha/2}k_i$ then we have $$\sum_i a_i \ge C_2,\;\;\sum_{i<j} (a_i-a_j)^2\le C_1 \epsilon^\alpha.$$
Now we are in good shape - decreasing $\epsilon$ will force the $a_i$ closer together without changing the lower bound. Letting $\hat a = \max_i a_i$, note that we have $$\hat a \ge \frac 1 n \sum_i^n a_i \ge \frac{C_2}n$$ and $$(a_i - \hat a)^2 \le \sum_{i<j}(a_i - a_j)^2\le C_1 \epsilon^\alpha;$$ so $$a_i \ge \hat a - |a_i-\hat a|\ge \frac{C_2}n-\sqrt{C_1 \epsilon^\alpha}.$$
Choosing $\epsilon$ small enough that $\sqrt{C_1 \epsilon^\alpha} < C_2/n$ completes the proof.