To solve the given optimization problem, we first need to translate the natural language description into a symbolic representation. This involves defining variables for each item (green beans and hamburgers), formulating the objective function using these variables, and then listing all constraints in terms of these variables.

Let's denote:
- \(x_1\) as the amount of green beans,
- \(x_2\) as the number of hamburgers.

The objective function to minimize is: \(8.4x_1 + 6.91x_2\).

Given constraints are:
1. Calcium from green beans and hamburgers must be at least 10 milligrams: \(5x_1 + 4x_2 \geq 10\).
2. The same constraint is repeated, so we keep it as is for clarity but note it's redundant: \(5x_1 + 4x_2 \geq 10\).
3. One times the number of green beans plus -1 times the number of hamburgers has to be no less than zero: \(x_1 - x_2 \geq 0\).
4. Total calcium from both sources must not exceed 21 milligrams: \(5x_1 + 4x_2 \leq 21\).

Since we can use fractional numbers of green beans and hamburgers, both \(x_1\) and \(x_2\) are continuous variables.

The symbolic representation in JSON format is:
```json
{
    'sym_variables': [('x1', 'green beans'), ('x2', 'hamburgers')],
    'objective_function': '8.4*x1 + 6.91*x2',
    'constraints': [
        '5*x1 + 4*x2 >= 10',
        'x1 - x2 >= 0',
        '5*x1 + 4*x2 <= 21'
    ]
}
```

Now, let's implement this optimization problem using Gurobi in Python:

```python
from gurobipy import *

# Create a new model
m = Model("Optimization_Problem")

# Add variables
x1 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="green_beans")
x2 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="hamburgers")

# Set the objective function
m.setObjective(8.4*x1 + 6.91*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(5*x1 + 4*x2 >= 10, "calcium_minimum")
m.addConstr(x1 - x2 >= 0, "green_beans_more_than_hamburgers")
m.addConstr(5*x1 + 4*x2 <= 21, "calcium_maximum")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Green beans: {x1.x}")
    print(f"Hamburgers: {x2.x}")
    print(f"Objective function value: {m.objVal}")
else:
    print("No optimal solution found")
```