We may assume that the $\lambda_i$ are distinct and that $\lambda_i \lt \lambda_j$ for $i \lt j$.
Given the $\lambda_i$ but no assumption on the $w_i,$ we can get the roots to be any set of $n-1$ points on the real line. In fact we can get it to be any set of $n-1$ or less distinct points except that the number is equal in parity to $n-1.$ So you need to be more specific. If the $w_i$ are all positive then the roots you seek interlace the $\lambda_i$ so the sum of the top $k$ roots is somewhere between $\sum_{n-k}^{n-1}\lambda_i$ and $\sum_{n-k+1}^{n}\lambda_i.$ These roots are also the places where $\prod|x-\lambda_i|^{m_i}$ has local maxima.
So perhaps for a reference you could look into interlacing of roots. However I don't have a good place for you to start.
The location of the roots is unchanged if we replace $p(x)$ with $cp(x)$ so we may assume $\sum w_i=1.$ The case $\sum w_i=0$ might be interesting , but I'll ignore it here.
The numerator of $$\sum_1^n\frac{w_i}{x-\lambda_i}=\frac{n(x)}{\prod(x-\lambda_i)}= \frac{x^{n-1}+\sum_0^{n-2}a_jx^j}{\prod(x-\lambda_i)} $$ is a monic polynomial of degree $n-1$.
Any choice of the $n-1$ values $w_1,w_2,\cdots,w_{n-1}$ (with $w_n=1-\sum_1^{n-1}w_i)$ uniquely determines the $n-1$ coefficients $a_0,a_1,\cdots,a_{n-2}.$ The converse is true as well, we can pick $n(x)$ to be any monic polynomial of degree $n-1$ and the $w_i$ are uniquely determined. It might be reasonable to specify that the $w_i$ are non-zero, but I won't in order to preserve this correspondence. From the way I specified the denominator, having a particular $w_i=0$ amounts to making $\lambda_i$ a root of $n(x).$
From now on I will assume that the $w_i$ are all positive and give two proofs that there is exactly one root in each interval $(\lambda_i,\lambda_{i+1})$
First proof: $p(x)$ is continuous in $(\lambda_{i},\lambda_{i+1}).$ Also, the one sided limits $$\lim_{x\to \lambda_i}p(x)=\pm\infty$$ with $+ \infty$ from above and $-\infty$ from below. This means the graph crosses the axis in each interval and there can only be $n-1$ roots.
Second proof: Consider $q(x)=\prod(|x-\lambda_i|)^{w_i}.$ It is continuous and non-negative touching the axis at each $\lambda_i$ but nowhere else. It has a local maximum in each $(\lambda_{i},\lambda_{i+1}).$ Also, $\ln |q(x)|$ has a local maximum at the same locations. Thus the derivative of $\ln|q(x)|$ has a zero in each interval (again at the same places). But that derivative is $p(x).$