To solve Adrian's problem, we first need to convert the natural language description into a symbolic representation of the optimization problem. Let's denote:

- $x_1$ as the number of bagels Adrian eats per day,
- $x_2$ as the number of burgers Adrian eats per day.

The objective function is to minimize costs. Given that each bagel costs $4.5 and each burger costs $12, the objective function can be represented algebraically as:

\[ \text{Minimize} \quad 4.5x_1 + 12x_2 \]

The constraints are:
1. Adrian wants to eat a minimum of 2500 calories per day. Since each bagel contains 250 calories and each burger contains 800 calories, this constraint can be represented as:

\[ 250x_1 + 800x_2 \geq 2500 \]

2. Adrian wants to eat at least 500 grams of fat per day. Given that each bagel contains 15 grams of fat and each burger contains 23.5 grams of fat, this constraint can be represented as:

\[ 15x_1 + 23.5x_2 \geq 500 \]

3. Since Adrian cannot eat a negative number of bagels or burgers, we also have non-negativity constraints:

\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Now, let's represent this problem in the requested format:

```json
{
    'sym_variables': [('x1', 'number of bagels'), ('x2', 'number of burgers')],
    'objective_function': '4.5*x1 + 12*x2',
    'constraints': [
        '250*x1 + 800*x2 >= 2500',
        '15*x1 + 23.5*x2 >= 500',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

Next, we'll write the Gurobi code to solve this linear programming problem:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="bagels")
x2 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="burgers")

# Set the objective function
m.setObjective(4.5*x1 + 12*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(250*x1 + 800*x2 >= 2500, "calories")
m.addConstr(15*x1 + 23.5*x2 >= 500, "fat")

# Optimize the model
m.optimize()

# Print the solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Bagels: {x1.x}")
    print(f"Burgers: {x2.x}")
    print(f"Total cost: ${4.5*x1.x + 12*x2.x:.2f}")
else:
    print("No optimal solution found")
```