1. We are given $f(x)=[x]-\left[\frac{x}{4}\right]$ and we need to examine the behavior at $x=4$.

2. Compute the value at $x=4$: $f(4)=[4]-\left[\frac{4}{4}\right]=4-[1]=4-1=3.$

3. Find the left-hand limit as $x\to 4^-$.

If $x\to 4^-$, then $x$ is slightly less than $4$, so $[x]=3$ and $\frac{x}{4}<1 \quad \Rightarrow \quad \left[\frac{x}{4}\right]=0.$ Hence, $f(x)=3-0=3.$ Therefore, $\lim_{x\to 4^-} f(x)=3.$

4. Find the right-hand limit as $x\to 4^+$.

If $x\to 4^+$, then $x$ is slightly greater than $4$ but less than $5$, so $[x]=4.$ Also, $1<\frac{x}{4}<\frac{5}{4} \quad \Rightarrow \quad \left[\frac{x}{4}\right]=1.$ Hence, $f(x)=4-1=3.$ Therefore, $\lim_{x\to 4^+} f(x)=3.$

5. Compare with $f(4)$: $\lim_{x\to 4^-}f(x)=\lim_{x\to 4^+}f(x)=3=f(4).$ So $f$ is continuous at $x=4$.

6. Check options:

• A: False, because both one-sided limits are equal.
• B: True, because $f$ is continuous at $x=4$.
• C: False.
• D: False.

Hence the correct option is $\boxed{\text{B}}.$