We need the points in $[0,8]$ where $f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$ is not continuous, and then sum those points.

1. Where can discontinuities occur?

A greatest integer function $[g(x)]$ is discontinuous exactly where $g(x)$ crosses an integer value.

So possible discontinuities of $\left[\frac{x}{2}+3\right]$ and $[\sqrt{x}]$ occur when:

1. $\dfrac{x}{2}+3 \in \mathbb Z$
2. $\sqrt{x} \in \mathbb Z$

within $x\in[0,8]$.

2. Discontinuities of $\left[\dfrac{x}{2}+3\right]$

We need $\frac{x}{2}+3=n, \quad n\in\mathbb Z$ so $x=2(n-3).$

Since $x\in[0,8]$, we get $\frac{x}{2}+3 \in [3,7].$ Thus possible integer values are $3,4,5,6,7$, giving $x=0,2,4,6,8.$

So $\left[\dfrac{x}{2}+3\right]$ is discontinuous at $x=0,2,4,6,8.$

3. Discontinuities of $[\sqrt{x}]$

We need $\sqrt{x}=m, \quad m\in\mathbb Z,$ so $x=m^2.$

For $x\in[0,8]$, possible squares are $x=0,1,4.$

So $[\sqrt{x}]$ is discontinuous at $x=0,1,4.$

4. Check discontinuities of the difference

The function is $f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}].$ A discontinuity may cancel if both terms jump by the same amount at the same point.

We check all candidate points: $x\in\{0,1,2,4,6,8\}.$

5. Test each point

(i) At $x=0$

Since domain is $[0,\infty)$, continuity at $0$ means right continuity.

For $x=0$: $f(0)=\left[3\right]-[0]=3.$ For small $x>0$: $\left[\frac{x}{2}+3\right]=3, \qquad [\sqrt{x}]=0,$ so $f(x)=3.$ Thus right limit equals $f(0)$, so $f$ is continuous at $0$.

(ii) At $x=1$

Near $x=1$:

• $\left[\dfrac{x}{2}+3\right]$ does not jump at $1$.
• $[\sqrt{x}]$ jumps from $0$ to $1$.

So $f$ jumps by $-1$, hence discontinuous at $1$.

(iii) At $x=2$

Near $x=2$:

• $\left[\dfrac{x}{2}+3\right]$ jumps from $3$ to $4$.
• $[\sqrt{x}]$ does not jump.

Hence $f$ jumps by $+1$, so discontinuous at $2$.

(iv) At $x=4$

Here both parts jump:

• $\left[\dfrac{x}{2}+3\right]$ jumps from $4$ to $5$.
• $[\sqrt{x}]$ jumps from $1$ to $2$.

Thus the change in $f$ is $+1-1=0.$ So discontinuity cancels.

Check explicitly:

• Just left of $4$: $f(x)=4-1=3$
• At/right of $4$: $f(x)=5-2=3$

So $f$ is continuous at $4$.

(v) At $x=6$

Near $x=6$:

• $\left[\dfrac{x}{2}+3\right]$ jumps from $5$ to $6$.
• $[\sqrt{x}]$ does not jump.

Hence $f$ is discontinuous at $6$.

(vi) At $x=8$

Near $x=8$:

• $\left[\dfrac{x}{2}+3\right]$ jumps from $6$ to $7$.
• $[\sqrt{x}]$ does not jump since $\sqrt{8}$ is not integer.

Hence $f$ is discontinuous at $8$.

6. Set of discontinuity points

Therefore, $S=\{1,2,6,8\}.$

So, $\sum_{a\in S} a = 1+2+6+8=17.$

7. Compare with stored answer

Stored correct answer: $17$

Our derived answer is also $17$, so they agree.