1. Given function

   $f(x)=\left[\frac{x^2}{2}\right]-[\sqrt{x}], \qquad x\in[0,4]$

   where $[\cdot]$ is the greatest integer function.

2. Idea

   A difference of two step functions is discontinuous at points where at least one of them jumps, unless the jumps cancel exactly.

   So we find the discontinuity points of:

   • $g(x)=\left[\frac{x^2}{2}\right]$
   • $h(x)=[\sqrt{x}]$

3. Discontinuities of $g(x)=\left[\frac{x^2}{2}\right]$

   Since $\dfrac{x^2}{2}$ is continuous and increasing on $[0,4]$, $g(x)$ changes value whenever $\frac{x^2}{2}=n$ for some integer $n$ in the range of $\dfrac{x^2}{2}$.

   Now, $\frac{x^2}{2}\in[0,8]$ so possible integers are $n=1,2,3,4,5,6,7,8.$

   Thus discontinuities occur at $x=\sqrt{2n}, \quad n=1,2,3,4,5,6,7,8.$

   These are: $\sqrt2,\;2,\;\sqrt6,\;2\sqrt2,\;\sqrt{10},\;\sqrt{12},\;\sqrt{14},\;4.$

4. Discontinuities of $h(x)=[\sqrt{x}]$

   Since $\sqrt{x}\in[0,2]$ for $x\in[0,4]$, $[\sqrt{x}]$ jumps when $\sqrt{x}=1 \quad \text{or} \quad \sqrt{x}=2.$

   So discontinuities are at $x=1,4.$

5. Combine the discontinuity points

   Candidate points are the union: $\{1,\sqrt2,2,\sqrt6,2\sqrt2,\sqrt{10},\sqrt{12},\sqrt{14},4\}$

   That is 9 points, but we must check whether any jump cancels.

6. Check each candidate point

   • At $x=1$: $\left[\frac{x^2}{2}\right]$ does not jump, since $\frac{1^2}{2}=\frac12$ is not an integer. $[\sqrt{x}]$ jumps from $0$ to $1$. Hence $f$ jumps. So $x=1$ is a discontinuity.

   • At $x=\sqrt{2n}$ for $n=1,2,3,4,5,6,7$: $\left[\frac{x^2}{2}\right]$ jumps by $1$. But $[\sqrt{x}]$ is continuous there unless $\sqrt{\sqrt{2n}}$ is an integer, which does not happen for these values. So all these are discontinuities.

   • At $x=2$ specifically: $\left[\frac{x^2}{2}\right]=\left[2\right]$ jumps, while $[\sqrt{x}]$ is continuous at $x=2$ since $\sqrt2$ is not an integer. So $x=2$ is a discontinuity.

   • At $x=4$: $\left[\frac{x^2}{2}\right]$ jumps because \frac{4^2}{2}=8,$$ and $$[\sqrt{x}]$$ also jumps because \sqrt4=2.$$

     Let us check the net jump:

     • just left of $4$: $\left[\frac{x^2}{2}\right]=7, \quad [\sqrt{x}]=1,$ so $f(4^-)=7-1=6.$
     • at $x=4$: $\left[\frac{x^2}{2}\right]=8, \quad [\sqrt{x}]=2,$ so $f(4)=8-2=6.$

     Hence the jumps cancel, so $f$ is continuous at $x=4$.

7. Total number of discontinuities

   We had 9 candidate points, but $x=4$ is not a discontinuity.

   Therefore the number of points of discontinuity is $9-1=8.$

Final Answer: $\boxed{8}$