For each i=1,.., n. There exist an integer k_i such that di=[k_i - bi, k_i -ai ] is a sub close interval of [0,1] The sum of those n segments is less than 1, so we could find a real number c that does not belong to any of di This number c satisfies ........

I dont quite understand your solution... Can some1 post another solution or can u help me understand yours? You say there exists $ k_i$ such that your $ d_i$ is in $ [0,1]$. But if thats the case, the task is solved,, we just let $ c=0$... Maybe im getting something wrong. My idea is something similar, but I didnt manage to complete it.

The author of this problem is prof. Vasile Pop, from Cluj. I'll post here the official solution.

Note that the problem is equivalent to finding some $c$ so that when we translate the intervals $(a_i,b_i)$ to $(a_i+c,b_i+c)$ none of the new intervals doesn't contain an integer. Now we can divide our intervals to intervals that already don't contain an integer and the ones that do.Let the first ones be called good and the others be bad. So let the first intervals $WLOG$ be $(c_i,d_i)$ such that $0<c_i<d_i<1$ and the others be $(1-a_i,1+b_i)$. Now note that all the bad intervals actually form one interval $(1-a,1+b)$ such that $a=max(a_1, \dots , a_n)$ and $b=max(b_1, \dots , b_n)$.Now note that we can rearrange good intervals to some new intervals $(c_i,d_i)$ such that they don't overlap. Now we can suposse that none of the good intervals overlaps with our bad interval(because if they do we can just make one bad interval out of those 2 intervals). Let our good intervals be $(c_1,d_1),(c_2,d_2), \dots ,(c_n,d_n)$ such that $c_1<d_1<c_2<d_2< \dots <c_n<d_n$. It also holds that $d_n<1-a$ so $d_n+a<1$ so we can choose some $\epsilon$ such that $d_n+a+\epsilon<1$ and $a+b+\epsilon<1$. Now by taking $c=a+\epsilon$ we see that our bad interval goes to good interval and that $(c_n,d_n)$ stays subset of $(0,1)$ so the other good intervals also stays subsets of $(0,1)$. We see that after translating for such chosen $c$ all of the intervals are good, so $c$ exists.

similar to official solution, just more expanded upon: Consider the quadratic polynomial $p_i(x)=(x+a_i)(x+b_i).$ We see that the roots of $p_i$ are $-a_i$ and $-b_i$, so that $p_i(c+k)\le 0$ iff $-b_i-c\le k\le -a_i-c$ for all integers $k$. Therefore, in order to show that there exists some $c\in\mathbb{R}$ such that $p_i(c+k)>0$ for all $i\in[n]$ and $k\in\mathbb{Z}$, it suffices to find some $c$ such that the interval $[-b_i-c,-a_i-c]$ and equivalently $$[a_i+c,b_i+c]$$contains no integers for all $i\in[n].$ Now, define $$S=\bigcup_{i=1}^n[a_i,b_i].$$Letting $l$ be the total length that $S$ encloses, we find that \begin{align*} l&=\sum_{i=1}^n(b_i-a_i)\\ &=\sum_{i=1}^nb_i-\sum_{i-1}^na_i\\ &<1, \end{align*}as given in the problem statement. Define the set $S'$ to contain the fractional parts of $S$, so that $x\in S$ iff $x-\lfloor x\rfloor\in S'.$ Clearly, $S'$ is a subset of $[0,1)$ and encloses a length of at most $l<1$, implying that $S'$ is a proper subset of $[0,1).$ Hence, there must exist some $c'\in[0,1)$ such that $c'\notin S'.$ But this means that $[a_i-c',b_i-c']$ cannot contain any integers for all $i\in[n]$, as otherwise there would be some $k'\in\mathbb{Z}$ such that $k'+c'\in[a_i,b_i]\subseteq S$, so that \begin{align*} c'&=(k'+c')-k'\\ &=(k'+c')-\lfloor k'+c'\rfloor\\ &\in S', \end{align*}a contradiction. Hence, setting $c'=-c$ implies the result, so we are done. $\blacksquare$