Let $T_1,\ldots,T_n>0$ be periods for $f_1,\ldots,f_n$ respectively. Now let's look at what happens to $f(x+p_1T_1+\ldots+p_nT_n)$ as some of the natural numbers $p_i$ go to $\infty$. Fix an $i$ and make just $p_i\to\infty$ while keeping all the other $p_j$'s constant. This will keep $f_i$ constant, and will show that $\sum_{j\ne i}f_j(x+\sum p_kT_k)$ has a limit when $p_i\to\infty$. Now increase some $p_j,\ j\ne i$ (for a single fixed $j\ne i$) with $1$, and then perform the same operation as above. $\sum_{j\ne i}f_j(x+\sum p_kT_k)$ will have the same limit that it had the last time, and this shows that $f(x+p_1T_1+\ldots+p_jT_j+\ldots+p_nT_n)=f(x+p_1T_1+\ldots+(p_j+1)T_j+\ldots+p_nT_n)$, which, in turn, means that $f_i$ has $T_j$ as a period. This means that all the $T_k$'s are periods for all the $f_k$'s, and from here the conclusion follows immediately. Sorry if it's a bit unclear, ehsan. Try it with $n=2$ to see what's going on.

A little bit clearer. By induction. For $n=1$ it is very simple. Assume it is true for $n-1$ and consider the function $ f(x+T_1)-f(x)$ that we know it has a limit 0. On the other hand, this function is the sum of $n-1$ periodic funtions $ f_i(x+T_1)-f_i(x)$ with $i>1$. Thus it must be constant and since it tends to 0, it must be 0. Thus $T_1$ is period for $f$ apply now the case $n=1$.

what a beautiful solutions thanks all of you