It's well-known that a ring $R$ is noetherian if and only if direct sums of injective $R$-modules are again injective.
How to prove the following characterization: $R$ is noetherian $\iff$ whenever $S_1,S_2,\ldots$ are simple right $R$-modules then $\bigoplus_{i=1}^{\infty}E(S_i)$ is injective as well, where $E(S_i)$ is the injective hull of $S_i$.
Assume that the direct sum of any countable collection of injective hulls of simple left $R$-modules is injective.
If $R$ is not left noetherian, there is an ascending chain of left ideals $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \cdots$ which not stops. Let $I = \bigcup_{n=1}^\infty I_n$, and note that for each $x \in I$ we have that $(x+I_n)_{n=1}^\infty \in \bigoplus_{n=1}^\infty I/I_n$, giving us a map $\tau \colon I \to \bigoplus_{n=1}^\infty I/I_n$.
Now, for each $n \geq 1$ we choose $x_n$ in $I \setminus I_n$, so that we can find a simple left $R$-module $S_n$ and a map $\sigma_n \colon I/I_n \to \text{E}(S_n)$ with $\sigma_n(x_n+I_n) \neq 0$. Let $\sigma$ be $\bigoplus_{n=1}^\infty \sigma_n \colon \bigoplus_{n=1}^\infty I/I_n \to \bigoplus_{n=1}^\infty \text{E}(S_n)$.
Then, $\sigma \circ \tau \colon I \to \bigoplus_{n=1}^\infty \text{E}(S_n)$ extends to a map $\rho \colon R \to \bigoplus_{n=1}^\infty \text{E}(S_n)$. Observe that, as $\rho(1)$ has almost all of its entries equal to zero, there exists $N \geq 1$ with the property that every element of $\rho(R)$ has a zero in each entry starting from the $N$-th one; a contradiction since the $N$-th entry of $\rho(x_N)$ is non-zero.