To solve the optimization problem described, we first need to translate the natural language description into a symbolic representation. This involves defining the variables, the objective function, and the constraints using algebraic terms.

### Symbolic Representation of the Problem

Given:
- Variables: `x1` for milligrams of vitamin B7, `x2` for milligrams of iron.
- Objective Function: Maximize `1.94*x1 + 6.5*x2`.
- Constraints:
  1. Kidney support index constraint: `13*x1 + 11*x2 >= 37`.
  2. Linear inequality constraint: `-10*x1 + 2*x2 >= 0`.
  3. Upper bound for kidney support index: `13*x1 + 11*x2 <= 95`.
  4. Non-negativity constraints are implicitly required but not explicitly mentioned, so we assume `x1 >= 0` and `x2 >= 0`.

The symbolic representation in JSON format is:

```json
{
  'sym_variables': [('x1', 'milligrams of vitamin B7'), ('x2', 'milligrams of iron')],
  'objective_function': 'Maximize 1.94*x1 + 6.5*x2',
  'constraints': [
    '13*x1 + 11*x2 >= 37',
    '-10*x1 + 2*x2 >= 0',
    '13*x1 + 11*x2 <= 95',
    'x1 >= 0',
    'x2 >= 0'
  ]
}
```

### Gurobi Code for the Optimization Problem

Now, we'll write the Gurobi code to solve this optimization problem. Note that `x2` must be an integer (non-fractional number of milligrams of iron), so we define it as an integer variable.

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="milligrams_of_vitamin_B7")
x2 = m.addVar(lb=0, vtype=GRB.INTEGER, name="milligrams_of_iron")

# Set the objective function
m.setObjective(1.94*x1 + 6.5*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(13*x1 + 11*x2 >= 37, name="kidney_support_index_lower_bound")
m.addConstr(-10*x1 + 2*x2 >= 0, name="linear_inequality_constraint")
m.addConstr(13*x1 + 11*x2 <= 95, name="kidney_support_index_upper_bound")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Milligrams of Vitamin B7: {x1.x}")
    print(f"Milligrams of Iron: {x2.x}")
    print(f"Objective Function Value: {m.objVal}")
else:
    print("No optimal solution found")
```