Step 1: Formulate the equations
Let $S, A,$ and $C$ be the number of students in Science, Arts, and Commerce respectively.
1) $S + A + C = 1500$
2) $1100S + 1000A + 800C = 15,50,000$

Step 2: Simplify the fee equation
Divide equation (2) by 100:
$11S + 10A + 8C = 15500$
Multiply equation (1) by 8:
$8S + 8A + 8C = 12000$
Subtract the two results:
$3S + 2A = 3500$

Step 3: Apply the constraint
The problem states: $S \le A$.
From $3S + 2A = 3500$, we can write $2A = 3500 - 3S$, so $A = \frac{3500 - 3S}{2}$.
Substitute this into the inequality:
$S \le \frac{3500 - 3S}{2}$
$2S \le 3500 - 3S$
$5S \le 3500 \implies S \le 700$.

Final Answer:
The maximum possible number of Science students is $\mathbf{700}$.