I am trying to minimize $(x_1+\dots+x_n)^2 + \frac{c}{2}\sum_{i=1}^n x_i^2 + \sum_{i=1}^n (\lambda_i - cz_i)x_i $ subject to $x_i\geq 0$ where $c\geq 1$ is a constant (that I can push arbitrarily large).
Clearly, if $\lambda_i-cz_i \geq 0$ then we can safely set $x_i^\ast=0$. So we can assume WLOG that $\lambda_i -cz_i <0$. Suppose $\lambda_1-cz_1 \leq \dots \leq \lambda_n- cz_n$. Then it also must be $x_1^\ast \geq \dots\geq x_n^\ast$:
Suppose not, then there is $x_i^\ast, x_j^\ast$ such that $\lambda_i-cz_i \leq \lambda_j - cz_j$ and $x_i^\ast<x_j^\ast$, but then changing $x_i^\ast, x_j^\ast$ with $x_i^\ast+\epsilon, x_j^\ast-\epsilon$ respectively makes a change of $\frac{c}{2}(x_i^\ast+\epsilon)^2+\frac{c}{2}(x_j^\ast-\epsilon)^2 -x_i^{2\ast} -x_j^{2\ast} + \epsilon(\lambda_i-cz_i - \lambda_j + cz_j) <0$ for sufficiently small $\epsilon$ which is a contradiction.
This seems to suggest an algorithm: Sort by $\lambda_i-cz_i$. Then for each $i=1, \dots, n$, suppose the first zero element is at $x_{i+1}=0$. Then try to only use $x_1, \dots, x_{i}$ in the solution. However, I'm stuck here. Any help?
There is probably some computation mistakes in the following, but I think there is some idea in it so I posted anyway. You are minimizing a convex function on a convex set, either the minimum of your function is in your domain in which case it is your solution, or you need to satisfy some of the $x_i\geq 0$ with equality. To know which one it is useful to actually know the minimum of your function with no constraint. We can differentiate the function with respect to $x_i$ to get $2(x_1+\dots+x_n)+c x_i+\lambda_i-cz_i$. Setting this to $0$ for all $i$ we get a linear system of equation, substracting the first one to the i'th one we get $c x_1+\lambda_1-c z_1-cx_i-\lambda_i+cz_i=0$ which is equivalent to $x_i=x_1+z_i-z_1+\frac{\lambda_1-\lambda_i}{c}$, which is also true for $i=1$. Pluggin this in the first equation we get $2(n(x_1-z_1)+z_1+\dots+z_n+\frac{n\lambda_1}{c}-\frac{\lambda_1+\dots+\lambda_n}{c})+cx_1+\lambda_1-cz_1=0$, solving for $x_1$ yields \begin{align*} x_1&=z_1-\frac{\lambda_1}{c}+2\frac{\lambda_1+\dots+\lambda_n}{c(2n+c)}-2\frac{(z_1+\dots+z_n)}{2n+c} \end{align*} and therefore \begin{align*} x_i&=z_i-\frac{\lambda_i}{c}+2\frac{\lambda_1+\dots+\lambda_n}{c(2n+c)}-2\frac{(z_1+\dots+z_n)}{2n+c}\\ &=\frac{1}{c}\left( c z_i-\lambda_i-\frac{2}{2n+c}\sum_{j=1}^n (cz_j-\lambda_j) \right) \end{align*}
If they are all positive then this is your answer, if for some $i$ it is negative, then you will (I'm not completely convince by this, it is probably true though) have $x_i=0$ and essentially minimize subject to those. But if $x_i=0$, then you can do the same optimization ignoring some subset of the indices, it will also remove elements from the solutions you have at the end. Let $v=\frac{2}{2n+c}\sum_{j=1}^n (cz_j-\lambda_j)$. If $\lambda_1-c z_1\geq \lambda_2-c z_2\geq\dots\geq \lambda_n-c z_n$ (sorry the order is not the same as yours) let $k$ be the smallest index such that $v\leq cz_k-\lambda_k$, then \begin{align*} x_i= \begin{cases} 0&\text{if } i<k\\ \frac{1}{c}\left( c z_i-\lambda_i-\frac{2}{2n+c}\sum_{j=1}^n (cz_j-\lambda_j) \right)&\text{if } k \leq i \end{cases} \end{align*}