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

Let's define:
- $x_1$ as the number of regular keyboards made each day,
- $x_2$ as the number of mechanical keyboards made each day.

The objective is to maximize profit. Given that the factory makes a profit of $30 per regular keyboard and $60 per mechanical keyboard, the objective function can be written as:
\[ \text{Maximize:} \quad 30x_1 + 60x_2 \]

Now, let's list the constraints based on the problem description:
1. Demand for regular keyboards: $x_1 \geq 165$
2. Demand for mechanical keyboards: $x_2 \geq 70$
3. Production limit for regular keyboards: $x_1 \leq 300$
4. Production limit for mechanical keyboards: $x_2 \leq 150$
5. Total keyboard production to satisfy the contract: $x_1 + x_2 \geq 250$

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'regular keyboards'), ('x2', 'mechanical keyboards')],
    'objective_function': '30*x1 + 60*x2',
    'constraints': [
        'x1 >= 165',
        'x2 >= 70',
        'x1 <= 300',
        'x2 <= 150',
        'x1 + x2 >= 250'
    ]
}
```

To solve this linear programming problem using Gurobi in Python, we will use the following code:

```python
from gurobipy import *

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

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

# Set objective function
m.setObjective(30*x1 + 60*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 >= 165, name="demand_regular")
m.addConstr(x2 >= 70, name="demand_mechanical")
m.addConstr(x1 <= 300, name="limit_regular")
m.addConstr(x2 <= 150, name="limit_mechanical")
m.addConstr(x1 + x2 >= 250, name="total_production")

# Optimize model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Regular keyboards: {x1.x}")
    print(f"Mechanical keyboards: {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```