Ratio of maximum sums in table You have a table with two rows and $k$ columns. The cells are filled with nonnegative real numbers so that the sum of the numbers in each row is $1$. 
Call the sum of the largest number in each column $X$ (so $X$ is a sum of $k$ numbers, one per column).
Color one number in each column, so that for each row, the sum of the colored numbers is at least the sum of the uncolored numbers, if we're allowed to remove an uncolored number of our choice. Call the maximum possible sum of the colored numbers $Y$.

What is the largest possible ratio $\frac{X}{Y}$?

On the one hand, the ratio cannot be larger than $1.5$. This is because if, in $X$, the sum of the colored numbers in each row is at least $0.5$, then $X=Y$. If not, $X\leq 1.5$, and we can check that $Y\geq 1$.
 A: 
What is the largest possible ratio $\frac{X}{Y}$?

Short answer: 8/7 based on the matrix
$$\begin{pmatrix}0.5 & 0.5 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{pmatrix}.$$
for which you get $X=4/3$ and $Y=7/6$ (by coloring the (1,1), (2,2) and (2,3) elements).
How I found this: consider the matrix $A\in\mathbb{R}_+^{2\times k}$ with elements $a_{ij}\geq 0$. We can always permute the columns in a way that the largest numbers are in the first row for the first set of columns, and in the second row for the last set of columns. Let $p$ denote the number of columns for which the maximum is in the first row. We will consider $p=1,2,\ldots,\left \lfloor{k/2}\right \rfloor$ (due to symmetry there is no need to consider larger $p$). We have:
$$X=\sum_{j=1}^p A_{1j} + \sum_{j=p+1}^k A_{2j}.$$
To further reduce symmetry, we can assume that the first $p$ columns and the last $k-p$ columns of $A$ are both sorted high to low in the first row. This symmetry breaking rule is not necessary, but helps in terms of speed (the first one is necessary for efficiently setting $X$).
Let $C \subset \{0,1\}^{2 \times k}$ denote the set of $2^k$ possible colorings: for $c\in C$, $c_{ij} = 1$ if element $(i,j)$ is colored, and 0 otherwise. The following inequalities set the highest lower bound on $Y$ by enumerating all valid colorings:
$$\forall c \in C : (Y \geq \sum_{ij} a_{ij} c_{ij}) \vee \sum_{j} a_{1j} c_{1j} < a_{1j} (1-c_{1j}) - a_{1j'} \forall j' \vee \sum_{j} a_{2j} c_{2j} < a_{2j} (1-c_{2j}) - a_{2j'} \forall j',$$
where $j'$ has to be an uncolored element $(c_{ij'}=0)$.
For fixed $k$, $p$ and $X$, you can minimize $Y$ via mixed integer optimization (matlab/YALMIP code below). It uses a big-M formulation for the three alternatives above. Here is the output for $k=4$:

If you run the code with $k$ larger than 4, you just get extra columns that are zero. Zooming in (generating more data points) shows a maximum at $X=4/3$.
k=5;
eps = 0.0001;
y = [];
for x = linspace(1,1.5,50)
    opt = 0;
    for p = 1:floor(k/2)
        yalmip('clear');
        A = sdpvar(2,k);
        Y = sdpvar(1,1);
        % nonnegative and row sum 1
        F = [A >= 0];
        F = F + [sum(A,2) == 1];
        % break symmetry
        F = F + [A(1,1:p) >= A(2,1:p)];
        F = F + [A(1,p+1:k) <= A(2,p+1:k)];
        F = F + [A(1,1:p-1) >= A(1,2:p)];
        F = F + [A(1,p:k-1) >= A(1,p+1:k)];
        % set X
        X = sum(A(1,1:p)) + sum(A(2,p+1:k));
        F = F + [X == x];
        % iterate over all colorings
        for c = (1 + dec2bin(1:2^k-2) - '0')'
            c = full(sparse(c,1:k,ones(k,1),2,k));
            % one out of three constraints
            one_out_of_three = binvar(3,1);
            F = F + [sum(one_out_of_three) == 1];
            F = F + [Y >= sum(sum(c.*A)) - k*(1-one_out_of_three(1))];
            sum_colored = sum(c.*A,2);
            sum_uncolored = sum((1-c).*A,2);
            for i = 1:2
                for j = find(c(i,:)==0)
                    F = F + [sum_colored(i) + 0.001 <= sum_uncolored(i) - A(i,j) + k*(1-one_out_of_three(1+i))];
                end
            end
        end
        optimize(F,Y, sdpsettings('verbose', 0));
        fprintf('Maximum ratio for X=%f, k=%d, p=%d: %f\n', x, k, p, double(X/Y));
        opt = max(opt, double(X/Y));
    end
    y = [y; opt];
end
plot(linspace(1,1.5,50), y, '*');
xlabel('X');
ylabel('max X/Y');

A: Before solving it, I will try to formalize this problem as an optimization problem
Calling the two rows by $R_1, R_2$ we have
$$
R_1 = \{r_1^1,r_1^2,\cdots,r_1^n\}\\
R_2 = \{r_2^1,r_2^2,\cdots,r_2^n\}
$$
such that $p_i(r_i) = r_i^j \ge 0$ and $e_i(r_i) = \sum_k r_i^k = 1$ for $i = \{1,2\}$
now 
$$
X(r_1,r_2) = \sum_k \max(r_1^k, r_2^k)
$$
considering now $\phi$ assuming values in $\{0,1\}$ let us define
$$
f(\phi,r_1,r_2) = \sum_k \phi_k r_1^k + \bar\phi_kr_2^k
$$
and the restrictions
$$
g_1(\phi,r_1) = \sum_k \phi_k r_1^k \ge \sum_k \bar\phi_k r_1^k\\
g_2(\phi,r_2) = \sum_k \phi_k r_2^k \ge \sum_k \bar\phi_k r_2^k
$$
and then with
$$
Y(\phi,r_1,r_2) = \max_{\phi} f(\phi,r_1,r_2)
$$
the final formulation
$$
\max_{\phi,r_1,r_2}\left(\frac XY\right) \ \ \mbox{s. t.}\ \ p_i(r_i)\cap e_i(r_i)\cap g_1(\phi,r_1)\cap g_2(\phi,r_2)
$$
