How many numbers can we select from $\{1,2,...2016\}$ such that sum of any four of them cannot be divided by $11$ How many numbers can we select from $\{1,2, \ldots, 2016\}$ such that sum of any four of them cannot be divided by $11$
It's not hard to come up with some combinations, but the question is how to prove it's the largest set. 
For example if we select numbers in forms of $11N+1,11N+4,11N+9$ will yield us $184 + 183 + 183$ numbers. 
it looks like proof will be somewhat of an inequalities problem. Let $a_{i_0}, a_{i_1}, \ldots, a_{i_k} > 0$ be the count of numbers we select from each modulo class, and we want to maximize $a_{i_0} + a_{i_1} + \cdots + a_{i_k}$ But how to express the constraint is tricky. 
 A: You can solve the problem via integer linear programming as follows.  For $j \in \{1,\dots,2016\}$, let binary decision variable $x_j$ indicate whether $j$ is selected.  The problem is to maximize $\sum_j x_j$ subject to linear constraints:
$$x_{j_1} + x_{j_2} + x_{j_3} + x_{j_4} \le 3$$
for all quadruples $j_1<j_2<j_3<j_4$ with $j_1+j_2+j_3+j_4 \equiv 0 \pmod{11}$.
That is too large to solve directly, but maybe you can reduce the problem size by aggregating variables from the same equivalence class.
Edit:
An alternative integer linear programming formulation uses binary variables $y_{k,c}$ to indicate whether equivalence class $k\in\{0,\dots,10\}$ has $c$ members selected.  The objective is to maximize $\sum_{k,c} c\cdot y_{k,c}$, and the constraints are 
$$\sum_c y_{k,c} = 1$$
for each $k$,
and 91 others of the form
$$\sum_{k,c} y_{k,c} \le b,$$
where $b \in \{0,1,2,3\}$.  For example, forbidden quadruple $(2,6,6,8)$ of equivalence classes corresponds to linear constraint
$$\sum_{c \ge 1} y_{2,c} + \sum_{c \ge 2} y_{6,c} + \sum_{c \ge 1} y_{8,c} \le 2.$$
Equivalently,
$$y_{2,0} + \sum_{c=0}^1 y_{6,c} + y_{8,0} \ge 1.$$
The resulting problem has 2027 variables and 102 constraints, and the optimal objective value is 552, in agreement with @dan_fulea.
