To solve the given optimization problem using Gurobi, we first need to understand and possibly simplify or clarify the constraints and objective function provided.

The objective function aims to maximize \(2.53 \times \text{number of packs of paper plates} + 1.37 \times \text{number of packs of napkins}\).

Given constraints:
1. The sustainability score from packs of paper plates is 7 each.
2. The sustainability score from packs of napkins is 9 each.
3. The total combined sustainability score must be at least 33.
4. \(10 \times \text{packs of paper plates} - 10 \times \text{packs of napkins} \geq 0\).
5. The total combined sustainability score must be no more than 47.
6. Packs of paper plates must be a non-fractional (integer) amount.
7. Packs of napkins must be an integer number.

Let's denote:
- \(x_0\) as the number of packs of paper plates,
- \(x_1\) as the number of packs of napkins.

The objective function can be written as: Maximize \(2.53x_0 + 1.37x_1\).

Constraints:
1. \(7x_0 + 9x_1 \geq 33\) (Minimum total sustainability score)
2. \(10x_0 - 10x_1 \geq 0\) (Which simplifies to \(x_0 \geq x_1\))
3. \(7x_0 + 9x_1 \leq 47\) (Maximum total sustainability score)
4. \(x_0, x_1 \in \mathbb{Z}^+\) (Both are non-negative integers, given the context of packs)

Now, let's write the Gurobi code to solve this problem:

```python
from gurobipy import *

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

# Define variables
x0 = m.addVar(vtype=GRB.INTEGER, name="packs_of_paper_plates")
x1 = m.addVar(vtype=GRB.INTEGER, name="packs_of_napkins")

# Set the objective function
m.setObjective(2.53*x0 + 1.37*x1, GRB.MAXIMIZE)

# Add constraints
m.addConstr(7*x0 + 9*x1 >= 33, "min_sustainability_score")
m.addConstr(10*x0 - 10*x1 >= 0, "balance_constraint")
m.addConstr(7*x0 + 9*x1 <= 47, "max_sustainability_score")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Packs of paper plates: {x0.x}")
    print(f"Packs of napkins: {x1.x}")
    print(f"Objective function value: {m.objVal}")
else:
    print("No optimal solution found")
```