To solve the given optimization problem, we need to define the variables and constraints according to the provided description. The objective function is to minimize $6x_0 + 2x_1$, where $x_0$ represents the milligrams of potassium and $x_1$ represents the milligrams of vitamin B2.

The constraints based on the immune support index and kidney support index for each nutrient are as follows:
- Immune support index from potassium: $26x_0$
- Immune support index from vitamin B2: $7x_1$
- Kidney support index from potassium: $25x_0$
- Kidney support index from vitamin B2: $21x_1$

The constraints are:
1. Total combined immune support index $\geq 19$: $26x_0 + 7x_1 \geq 19$
2. Total combined kidney support index $\geq 22$: $25x_0 + 21x_1 \geq 22$
3. The condition $-6x_0 + 6x_1 \geq 0$ simplifies to $x_1 \geq x_0$.
4. Total combined immune support index $\leq 48$: $26x_0 + 7x_1 \leq 48$
5. Total combined kidney support index $\leq 45$: $25x_0 + 21x_1 \leq 45$

Since both $x_0$ and $x_1$ can be non-integer, we don't need to specify them as integers in the model.

Here is how you could set up this problem using Gurobi in Python:

```python
from gurobipy import *

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

# Define variables
x0 = m.addVar(name='milligrams_of_potassium', lb=0, ub=GRB.INFINITY)
x1 = m.addVar(name='milligrams_of_vitamin_B2', lb=0, ub=GRB.INFINITY)

# Objective function: Minimize 6*x0 + 2*x1
m.setObjective(6*x0 + 2*x1, GRB.MINIMIZE)

# Constraints
m.addConstr(26*x0 + 7*x1 >= 19, name='immune_support_index_min')
m.addConstr(25*x0 + 21*x1 >= 22, name='kidney_support_index_min')
m.addConstr(x1 >= x0, name='vitamin_b2_vs_potassium')
m.addConstr(26*x0 + 7*x1 <= 48, name='immune_support_index_max')
m.addConstr(25*x0 + 21*x1 <= 45, name='kidney_support_index_max')

# Solve the model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Milligrams of Potassium: {x0.x}")
    print(f"Milligrams of Vitamin B2: {x1.x}")
    print(f"Objective Function Value: {m.objVal}")
else:
    print("No optimal solution found")
```