This is mainly a comment. My first guess would be that if it is true, then a random subset of density $1/2$ works. A result in this direction is Lemma 3.2 in 

M. Rudelson, *Contact points of convex bodies*, Israel Journal of Mathematics, 1997, Volume 101, Number 1, Pages 93-124

says:

> **Lemma** :Let $x_1,...,x_k$ be vectors in $\mathbb R^n$, $\varepsilon_1,...,\varepsilon_k$ be independent Bernoulli variables, taking values $1,-1$ with probability $1/2$. Then
$${\mathbb E} \left\| \sum_{i=1}^k \varepsilon_i |x_i\rangle {\langle x_i} | \right\|\leq C \log(n) \sqrt{\log(k)} \max_i \|x_i\| \left\| \sum_{i=1}^k |x_i\rangle \langle x_i| \right\|^{1/2}$$
for some absolute constant C.


In your case, this says that a random set of density $1/2$ solves the easier problem where $\delta$ may depend on $n$ and $k$. At the same time it proves much more, since there is even concentration around $1/2$. More precisely, if $ \sum_{i=1}^k |x_i\rangle \langle x_i| =1$ and $\|x_i\|< \delta$, then

$${\mathbb E} \left\|\frac12 - \sum_{i=1}^k \eta_i |x_i\rangle {\langle x_i} | \right\|\leq C/2 \log(n) \sqrt{\log(k)} \cdot \delta,$$
where $\eta_i$ are independent Bernoulli with values in $\{0,1\}$. 

Since $n \leq k \delta^2$ (looking at the trace), an easy calculation shows that $\delta$ only depends on $k$. At the same time, it seems to me that the problem is getting easier if $k$ is larger, but I cannot substantiate this claim. Remark 3.3 in the same paper shows that the inequality cannot be improved to become independent of $n$ and $k$. This somehow shows that choosing a random subset is too naive; at least when one studies the expected value of the norm as in the inequality above.