Let $A$ be a graded algebra over a commutative ring, and $I$ a two sided ideal of $A$ which is not necessarily graded. We can take the quotient $A/I$ and obtain an algebra, however I am struggling to see why we need $I$ to be graded for $A/I$ to be graded.
For example, let $a\in A$ be arbitrary, then $a$ can be written as the finite sum:
$$a=\sum_na_n$$
where each $a_n$ is homogenous, i.e. $a_n\in A_n$. Since the projection is an algebra homomorphism, we have that:
$$[a]=\sum_n[a_n]$$
so each element $[a]\in A/I$ can be written as a finite sum over $n$, where each $[a_n]\in \pi(A_n)$, i.e. the restriction of $\pi$ to the subspace $A_n$. For brevity denote $\pi(A_n)$ by $(A/I)_n$, then it the above implies that:
$$(A/I)=\bigoplus_{n=0}^\infty (A/I)_n$$
By the induced multiplicative structure on $A/I$ we have that for $[a_n]\in (A/I)_n$ and $[a_m]\in (A/I)_m$: $$[a_n]\cdot[a_m]=[a_n\cdot a_m]\in \pi(A_{n+m})=(A/I)_{m+n}$$ so $(A/I)_n(A/I)_m\subset (A/I)_{m+n}$. Does this not imply that $(A/I)$ is graded? I suspect I have made some blunder somewhere, but I can't see it. I do see how we need $I$ to be graded to write:
$$A/I=\bigoplus_{n=0}^\infty A_n/I_n$$
but I'm not sure why we need such a stringent condition for $A$ to admit some form of grading.
Edit:
Let $I$ be graded, and suppose that $[a]\in \pi(A_n)\cap \pi(A_m)$ for some $n\neq m$. It then follows that there exists an $a_n\in A_n$ and $a_m\in A_m$ such that: $$[a_n]=[a]=[a_m]$$ implying that for some element $i\in I$:
$$a_n=a_m+i$$
Since $I$ is graded, we can decompose $i$ into a summation over homogenous elements $b_i\in A_i\cap I$, hence:
$$a_n=a_m+\sum_{i}b_i$$
This implies that: $$\sum_{i}b_i=a_n-a_m$$ and since $i$ is a sum over homogenous elements, we have that $b_i=0$ for $i\neq n,m$, $b_n=a_n$, and $b_m=a_m$. This then implies that $a_n\in I\cap A_n$ and $a_m\in I\cap A_m$, hence: $$[a_n]=[a_m]=0$$ so $\pi(A_n)\cap \pi(A_m)=\{0\}$ as desired.
Does this work?