Step 1: Set up the Equation
Let $B$ be the initial number of boys and $G$ be the initial number of girls.
Remaining girls = $0.6G$. Remaining boys = $0.4B$.
Given: $0.6G = 0.4B + 8$. Multiply by 10: $6G = 4B + 80 \implies 3G = 2B + 40$.

Step 2: Apply Constraints
1. $B$ and $G$ must be integers.
2. $0.4B$ and $0.6G$ must be integers (since people are discrete). This means $B$ is a multiple of 5 and $G$ is a multiple of 5.
3. $B > 10$.

Step 3: Test values for B (multiples of 5)
- If $B = 15$: $3G = 2(15) + 40 = 70$ (Not divisible by 3).
- If $B = 20$: $3G = 2(20) + 40 = 80$ (Not divisible by 3).
- If $B = 25$: $3G = 2(25) + 40 = 90 \implies G = 30$.
Total initial students = $B + G = 25 + 30 = \mathbf{55}$.