Let $n$ be a positive integer. Find the non-negative integer $m$ that satisfies $\sum_{k=0}^{m}\left[\frac{n-\log _{2}(2 k+1)}{2}\right]=0$, where $[x]$ denotes the greatest integer less than or equal to $x$.

---

The translation preserves the original text's formatting and structure.
Solution: Let the positive integer $k_{0}$ satisfy
$$
\begin{array}{l}
\frac{n-\log _{2}\left(2 k_{0}+1\right)}{2} \geqslant 0, \\
\frac{n-\log _{2}\left[2\left(k_{0}+1\right)+1\right]}{2}1$ when, $p(n)>0$, so it is necessary to make
$$
\sum_{k=2^{n-1}}^{m}\left[\frac{n-\log _{2}(2 k+1)}{2}\right]=-\frac{2^{n-1}+(-1)^{n}}{3} .
$$

Similarly, for $k=2^{n-1}, 2^{n-1}+1,2^{n-1}+2, \cdots$, $2^{n+1}-1$ we have
$$
\left[\frac{n-\log _{2}(2 k+1)}{2}\right]=-1 \text {. }
$$

Therefore, it is only necessary to let $m=\frac{2^{n+1}+(-1)^{n}-3}{3}$.
And $2^{n-1} \leqslant m=\frac{2^{n+1}+(-1)^{n}-3}{3}1$ when,
$$
m=\frac{2^{n+1}+(-1)^{n}-3}{3} \text {. }
$$

When $n=1$, it is easy to know that $m=0$, which satisfies the expression for $m$.
Therefore, $m=\frac{2^{n+1}+(-1)^{n}-3}{3}$.
(Su Weijie, Class 13, Grade 2, Yuyao High School, Yuyao City, Zhejiang Province, 315400)