*[update] By an example of 4x4-matrices the ansatz below could not be used to solve the problem. The matrix* $\small Q_K $ *cannot in general be made lower triangular by choices of the* $\small k_i $. *I'll delete this answer soon if I cannot improve the ansatz*.

I do not yet have a full answer but possibly a first step into one. I think that this can be answered unsing two facts:

a) There is a similarity transformation with a rotation *T* such that $\small P = T^{-1}\cdot A \cdot T = T' \cdot A \cdot T $ where *P* is triangular and has the eigenvalues of *A* on its diagonal.

b) Your sum-expression of vector outer products, let it be called matrix $\small E_K = \sum_{i=1}^n B_i k_i C_i = \sum_{i=1}^n k_i (B_i C_i)= \sum_{i=1}^n k_i E_i $ is a weighted (by the $\small k_i $ weights) sum of rank-*1*-matrices $\small E_i $

From the "similarity rotated version" of all matrices

$\qquad \small Q_K = T' E_K T = \sum_{i=1}^n k_i (T' E_i T) = \sum_{i=1}^n k_i Q_i $ (which should be made triangular by choices of $\small k_i $ ) and

$\qquad \small R = T' D T $ which is then also triangular

we get your final equation in its form with triangular matrices

$\qquad \small R = P + Q_K $

We'll have a solution if the weights $\small k_i $ for the *non-triangular*, generic but *rank-1*-matrices $\small Q_i $ can be chosen such that their sum $\small Q_K$ becomes triangular *and* its diagonal equals the negative diagonal in $\small P $.

I've a vague speculation that the equation with this triangular matrices can easier be shown to be "almost always" possible, but have not yet a further concrete approach. At least this construction exhibits that the rank-1-matrices $\small E_i $ (and thus $\small Q_i$ ) cannot be simple scalar multiples of each other if *A* has full rank and thus the restriction to "for almost all" is unavoidable and possibly mainly consists of this property.