Let $R$ be a ring in which every maximal ideal is a direct sum of cyclic $R$-modules. Now let $I$ be a proper ideal of $R$. What is the structure of $I$? Is it true that $I$ is a direct sum of cyclic $R$-modules? If not, what is a counter-example?
Note that, we also know that, the above is true for a Noetherian local ring, i.e., if $R$ is a local Noetherian ring and $M=Rx\oplus Ry$, then every ideal of $R$ is a direct sum of at most two cyclic $R$-modules.
Let $R$ be the ring with the property you desired. Let $I\subset J\subset R$ and $J$ be the maximal ideal. Then $I$ is an $R$-submodule of $J$. By assumption we have $$ J=\oplus_{l\in L}Rj_{i}=\oplus_{l\in L} J_{l} $$ Let $l\in L$ be any given element in the index set. Let $I\cap Rj_{l}=H_{l}$. If $R$ is Noetherian, then $H_{l}$ is finitely generated. So we can set up an inductive procedure by picking up elements $h_{k}$ in $H$ and check if $H=\oplus^{n}_{k=1}Rh_{k}$ at every step. Since this must stop somewhere, we may conclude that $R=\oplus_{l\in L, k\le n_{l}}Rh_{k}^{l}$. However it is not clear to me if the same procedure carries over if $R$ is not Noetherian.