Assume we have $k$ dependent random variables $X_{1}, \dots, X_{k}$ with $\operatorname{Var}(X_{i}) < \infty$.
In which case
$$\sqrt{k}\operatorname{Var}\left(\sum_{i=1}^{k}X_{i}\right) \leq \sum_{i=1}^{k}\operatorname{Var}(X_{i})\,? $$
It seems that negative covariance is not enough.
Starting with the identity $$\operatorname{Var} \left( \sum_{i=1}^k X_i \right) = \sum_{i=1}^k \operatorname{Var}(X_i) + \sum_{i > j} 2\operatorname{Cov}(X_i, X_j)$$ we can see why having negative covariance isn't sufficient; we need the covariances to be negative enough to offset the $\sqrt k$ multiplier. Specifically, what we need is
\begin{align*} & \sqrt k\sum_{i=1}^k \operatorname{Var}(X_i) + \sqrt k \sum_{i > j} 2\operatorname{Cov}(X_i, X_j) \leq \sum_{i=1}^k \operatorname{Var}(X_i) \\ \iff & \sum_{i > j} \operatorname{Cov}(X_i, X_j) \leq \frac{k^{-1/2} - 1}{2 } \sum_{i=1}^k \operatorname{Var}(X_i) \end{align*} and we note that the coefficient on the right is always negative when $k \geq 2$, implying that having negative total covariances is necessary but not sufficient.