What is your idea about this conjecture?

I conjecture that in a consecutive sequence of $n$ natural numbers all greater than $n$, there exists at least one number which is not divisible by any prime number less than or equal to $n/2$.

Can any one prove or disprove this?

The conjecture is false. This problem is directly related to finding large gaps between primes, and the methods of Erdos, Rankin and others.

Define the Jacobsthal function $j(q)$ to be the largest gap between consecutive reduced residues modulo $q$, that is the largest gap between elements that are relatively prime to $q$. Note that your conjecture is equivalent to asking if $$j\left(\prod_{p\leq n/2} p\right)\leq n$$ holds for all $n$. To see why, consider any sequence of $n$ consecutive numbers modulo $M=\prod_{p\leq n/2}p$. Then each of them will be divisible by some $p\leq n/2$ if and only if $j\left(\prod_{p\leq n/2}p\right)\geq n$.

This function $j$ is directly related to best lower bounds for prime gaps. Indeed, if $$j\left(\prod_{p\leq X}p\right)\geq f(X)$$ infinitely often, (where $f$ is a nice function, strictly increasing etc.) then $$\max_{p_{n+1}\leq x} p_{n+1}-p_n \geq f(\log x).$$ In a recent paper of Kevin Ford, Ben Green, Sergei Konyagin, James Maynard, Terence Tao, they proved that $$j\left(\prod_{p\leq x} p\right)\gg \frac{x\log x \log \log \log x}{\log \log x},$$ and hence

$$\max_{p_n\leq X} p_{n+1}-p_n\gg \frac{\log X\log \log X \log \log \log \log X}{\log \log \log X}.$$ We remark that Erdos had put a $10000\$$ prize on this result, the largest amount he set for any problem.

While this result does disprove your conjecture, we do not need use such powerful theorems. We need only use Lemma 7.13 of Montgomery and Vaughn which states that $$\lim_{n\rightarrow \infty} \frac{j\left(\prod_{p\leq n} p\right)}{n}=\infty.$$ This is proven using an elementary sieving argument, and the result was originally given by Westzynthius.