is the problem has no solution?????

Any solution ?

Randomly select a subset of $\{ 1,2,\dotsc ,2m\}$ and let $X$ be the number of its element. Easy to compute that the expected value of $X$ is $m$ and variance $\sigma^2$ is equal to $m/2$. By Chebychev's Inequality, we get that for all $k>0$, $$\Pr \left( |X-m|\geq k\sqrt{m/2}\right) \leqslant \frac{1}{k^2}.$$Plugging $k$ by $(t+1)\sqrt{2/m}$ gives us $$\frac{2\left( \dbinom{2m}{0}+\dbinom{2m}{1}+\cdots +\dbinom{2m}{m-t-1}\right) }{2^{2m}} \leqslant \frac{m}{2(t+1)^2}<\frac{1}{2k}.$$This gives the stronger inequality than desired.