Consider the following linear program - and then you write down a bilinear program.
Good news though is that you easily can solve this by bisection in $y$, meaning that
complexity is complexity of solving a linear program, times a constant which is logarithmic in the desired precision (as you half the interval for every linear program you solve)
EDIT: Good or bad news is that the problem can be solved analytically, the optimal value approaches $0$ but that is done by letting half of the $x_i$ tend to infinity. Let $x_i$ be $\mu$ for even $i$ and $0$ for odd $i$. The constraints involving $z$ then all simplify to $z\leq \mu$. The constraints will be active at optimality (otherwise $z$ can be made larger and thus $y$ smaller), hence $z = \mu$. The first constraint simplifies to $(c_1\mu \lfloor n/2 \rfloor -c_2)/\mu\leq y$, or equivalently we are minimizing $c_1 \lfloor n/2 \rfloor -c_2/\mu$. We have $\mu \geq 0$ and for the problem to make sense $c_2\leq 0$ (otherwise the problem is trivial as one could take the zero solution.)
Hence, the infimum is $c_1 \lfloor n/2 \rfloor$ which is approached when $\mu \rightarrow 0$
My interpretation is that the question is: How can I assure that the linear inequality $q_{ij}-q_{il}-q_{kj}+q_{kl} \geq 0$ holds when $q_{ij} \geq q_{kl}$.
This is an implication betweeen linear inequalities
$$q_{ij} \geq q_{kl} \Rightarrow q_{ij}-q_{il}-q_{kj}+q_{kl} \geq 0 $$
Unfortunately not LP-representable. However, it can be modelled using auxilliary binary variables (thus leading to a mixed-integer quadratic program)
For notational simplicty, consider the generic case for some constraints (not even necessarily linear) $p(x)\geq 0$ and $w(x)\geq 0$ and decision variable $x$
$$p(x) \geq 0 \Rightarrow w(x) \geq 0 $$
Introduce a binary variable $\delta$ and a large number $M$ (more later). The implication is guaranteed to hold if the following two constraints hold
$$p(x) \leq M \delta, w(x)\geq -M(1-\delta)$$
If $p(x)$ is positive, $\delta$ will be forced to be $1$, which forces $w(x)$ to be non-negative.
This is called big-M modelling. When you implement, $M$ should not be made unnecessarily large, but just large enough so that it doesn't affect the feasible space in $x$ when the binary variable is inactive. In your case, assuming the expressions are built from $q$ the difference and sums between your up to 4 $q$ variables can trivially never exceed 4, hence $M=4$ or something like that should be possible. If the variables $a$ and $b$ are involved, you have to know bounds on these so you can bound the expressions accordingly.
If $p(x)$ is a vector of length $n$, and you want the condition to be activated when all elements are non-negative, you would introduce a vector binary variable $\Delta$ in addition to your previous $\delta$, and use $p(x) \leq M\Delta, \delta \geq 1 + \sum\Delta - n$. If all elements are positive, all $\Delta$ binaries will be forced to be $1$, forcing $\delta$ to be $1$.
Best Answer
Convex constraints in standard form are given by $$g(\mathbf{x}) \le 0,$$ where $g(\cdot)$ is a convex function. Your constraints are already in the form given above, but your $g(\cdot)$ functions are not convex.
To check whether $g(\cdot)$ is convex, you need to check whether the Hessian -- the matrix of partial second derivatives -- is positive semidefinite. So, for example, the Hessian of your first constraint is $$\left[\matrix{0 & 0 & 1.4 & 0 \\ 0 & 0 & 0 & -1 \\ 1.4 & 0 & 0 & 0 \\0 & -1 & 0 & 0}\right],$$ which is not PSD (it has negative eigenvalues $\implies$ not PSD).
I gather that you are trying to find a way to "write" a nonconvex constraint as a convex one. But it's not a matter of just writing it in a specific way. A constraint is either convex or not; shifting things to one side of the inequality or the other, multiplying both sides by $-1$, etc. --- things like that are not going to convert a nonconvex constraint into a convex one.