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

Let's denote:
- $x_1$ as the amount invested in clean water,
- $x_2$ as the amount invested in electricity.

The objective is to maximize total profit. Given that each dollar invested in clean water yields a $1.9 profit and each dollar invested in electricity yields a $2.3 profit, the objective function can be represented as:
\[ \text{Maximize: } 1.9x_1 + 2.3x_2 \]

The constraints based on the problem description are:
1. The total budget is up to $5500, so \( x_1 + x_2 \leq 5500 \).
2. No less than $1000 must be invested in clean water, so \( x_1 \geq 1000 \).
3. No less than 30% of all money invested must be in electricity. This can be represented as \( x_2 \geq 0.3(x_1 + x_2) \), which simplifies to \( 0.7x_2 \geq 0.3x_1 \) or \( 7x_2 \geq 3x_1 \).

Additionally, since we are dealing with investments, both $x_1$ and $x_2$ should be non-negative (\( x_1 \geq 0 \), \( x_2 \geq 0 \)).

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'amount invested in clean water'), ('x2', 'amount invested in electricity')],
    'objective_function': 'Maximize: 1.9*x1 + 2.3*x2',
    'constraints': ['x1 + x2 <= 5500', 'x1 >= 1000', '7*x2 >= 3*x1', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we can use the following code:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, name="clean_water")
x2 = m.addVar(lb=0, name="electricity")

# Set the objective function
m.setObjective(1.9*x1 + 2.3*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + x2 <= 5500, "budget")
m.addConstr(x1 >= 1000, "clean_water_min")
m.addConstr(7*x2 >= 3*x1, "electricity_percentage")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Amount invested in clean water: {x1.x}")
    print(f"Amount invested in electricity: {x2.x}")
    print(f"Total profit: {m.objVal}")
else:
    print("No optimal solution found")
```