Let $R$ be a ring with identity and $A_1,\dots,A_n$ ideals of $R$ with the property, that $A_i + A_j = R$ whenever $i \neq j$. Then $$R = A_i + \bigcap_{j \neq i} A_j \qquad i = 1,\dots,n$$
The easiest way is propably to show that $1$ belongs to the righthandside. Fix some $i$. Then we find $a_j \in A_i$ and $b_j \in A_j$ for $j \neq i$ such that $$1 = a_j + b_j$$ Hence $$1 = (a_1 + b_1)\dots(a_n + b_n)$$ Now I do not know how to conclude.
EDIT: This solution is not right. See the comment below and the other answer.
NEW EDIT: Thanks to @user26857 this answer has been fixed and now is certainly right.
I think your idea is right. It's enough to prove that $1\in A_i+\bigcap_{j\neq i} A_j$. For a fixed $i$ we have by hypothesis, $1 = x_{i,j}+x_j$ for every $j\neq i$, so $$1=(x_{i,1}+x_1)\cdots (x_{i,i-1}+x_{i-1})(x_{i,i+1}+x_{i+1})\cdots (x_{i,n}+x_n)=(\text{something from }I_i)+x_1\cdots \hat{x_i}\cdots x_n.$$
(The symbol $^\hat{}$ means that we don't consider that term). Now, since the $A_j$'s are ideals, we deduce that $$x_1\cdots \hat{x_i}\cdots x_n\in \bigcap_{j\neq i} A_j$$ $$\implies (\text{something from }I_i)+x_1\cdots \hat{x_i}\cdots x_n\in A_i+\bigcap_{j\neq i} A_j.$$
Therefore $1\in A_i+\bigcap_{j\neq i} A_j$ and hence $$R=A_i+\bigcap_{j\neq i} A_j\qquad \forall \,i=1,\dots, n.$$
I have a little bit concern about Xam's answer. Indeed for a fixed $i$ and every $j$ we have $1=x_i+x_j$ for $x_i\in A_i$ and $x_j\in A_j$. But for a different $j'\ne j$, the $x_i'$ in $1=x_i'+x_{j'}$ will probably be different with $x_i$.
I can't write a comment there so I can only write a new proof below
For every $j\ne i$, $1=x_j+y_j$ for $x_j\in A_i$ and $y_j\in A_j$, so $$1=\prod_{j\ne i}(x_j+y_j)=f(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n,y_1,\ldots,y_{i-1},y_{i+1},\ldots,y_n)+\prod_{j\ne i}y_j$$ where every term in $f$ has a $x_j$ with positive degree. So each term of $f$ is in $A_i$, $f$ is in $A_i$. And $\prod_{j\ne i}y_j\in\cap_{j\ne i}A_j$. Therefore $1\in A_i+\cap_{j\ne i}A_j$ and $R\in A_i+\cap_{j\ne i}A_j$.