To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables and formulating the objective function and constraints.

Let's define:
- \(x_1\) as the number of zippered hoodies.
- \(x_2\) as the number of pullover hoodies.

The objective function, which we want to maximize, is the total profit from selling both types of hoodies. Given that the profit per zippered hoodie is $5 and the profit per pullover hoodie is $4, the objective function can be represented algebraically as:
\[ \text{Maximize:} \quad 5x_1 + 4x_2 \]

The constraints are:
1. The total cost constraint: The boy can spend a total of $450. Since zippered hoodies cost $20 each and pullover hoodies cost $15 each, we have:
\[ 20x_1 + 15x_2 \leq 450 \]
2. The total number of hoodies constraint: He can sell at most 25 hoodies total, which gives us:
\[ x_1 + x_2 \leq 25 \]
3. Non-negativity constraints: Since the number of hoodies cannot be negative, we also have:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Symbolic representation in JSON format:
```json
{
    'sym_variables': [('x1', 'number of zippered hoodies'), ('x2', 'number of pullover hoodies')],
    'objective_function': '5*x1 + 4*x2',
    'constraints': ['20*x1 + 15*x2 <= 450', 'x1 + x2 <= 25', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.INTEGER, name="zippered_hoodies")
x2 = m.addVar(lb=0, vtype=GRB.INTEGER, name="pullover_hoodies")

# Set the objective function
m.setObjective(5*x1 + 4*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(20*x1 + 15*x2 <= 450, "total_cost")
m.addConstr(x1 + x2 <= 25, "total_hoodies")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of zippered hoodies: {x1.x}")
    print(f"Number of pullover hoodies: {x2.x}")
    print(f"Total profit: {m.objVal}")
else:
    print("No optimal solution found")
```