We must assume that $\langle k \rangle \ne 0.$ When $n=1$ we know everything.
Certainly $\langle k^2\rangle \ge\langle k\rangle ^2.$ If $\langle k^2\rangle =\langle k\rangle ^2$ then the $k_i$ are all equal (to $\langle k \rangle$ ) and $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}=\langle x \rangle$ exactly. The same thing happens if all the $x_i$ are equal although we have no way of know from the given information if that is the case. Also knowing $\sum_{i=1}^{n} x_i^2$ would be helpful. Then $\sum_{i=1}^{n} k_i x_i=C\sqrt{\sum_{i=1}^{n} k_i^2\sum_{i=1}^{n} x_i^2}$ for some $0 \le C \le 1.$
In the special case that $n=3$, and $\langle k^2 \rangle=30$ we can figure that the $k_i$ are $1,2,5$ in some order (given that they are integers). I will ignore these number theoretic features and only assume that the $k_i$ are non-negative reals.
We may assume that $\max_i{x_i}=x_1.$ When $\langle k^2 \rangle \gt \langle k \rangle^2$ (i.e. the $k_i$ are not all equal) all we can say with the given information is that $0 \le \frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i} \le x_1.$ Equality occurs when $k_1 \gt 0$ but $k_i=0$ for $i \gt 1.$ That only one of the $k_i$ is non-zero can be discovered from $\langle k^2 \rangle =n\langle k \rangle^2$ Now $\max_i{x_i} \le \sum_{i=1}^{n} x_i=n\langle x \rangle$ with equality when $x_i=0$ for $i \ge 2.$
Earlier Careless reading lead me to mention: If the $k_i$ are not all equal, and some of the $x_i$ can be negative, then nothing can be deduced. even if we actually know all of the $k_i.$ Let us assume that $k_2=k_1+\epsilon.$ Assume first that $n=2$. Then given desired values $\langle x \rangle=m$ and $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}=S$ we must take $x_1=m+\delta$ and $x_2=m-\delta$ for $\delta=\frac{(S-m)(2k_1+\epsilon)}{\epsilon}.$ For larger $n$ I can volunteer the extra information that $x_i=0$ for all $i \gt 2$ and adjust the formulas slightly.
