Let us first compute the electric field on the surface of a spherical (infinitely thin) shell with a uniform charge distribution. We can use Gauss's law for this, $$\int_{S_r} \vec{E}\cdot d\vec{A} = \frac{Q_{in}}{\varepsilon_0},$$ from which using that the charge density is $\sigma = Q/4\pi R^2$ and taking the Gauss surface at $r=R$, we get (using the spherical symmetry): $$E = \frac{4\pi R^2\sigma}{4\pi r^2\varepsilon_0} \stackrel{r\longrightarrow R}{=} k_e\frac{Q}{R^2}$$ as you stated yourself. However notice that for all things to hold, the limit is taken from above, namely the Gaussian surface is taken from outside and reduced. On the other hand we also know the field inside will be zero, so there is a clear discontinuity in the electric field at $r=R$. So formally the field is not defined there, but we regularize the field by taking the average of the discontinuity jump. In this case, $$E(R):= \frac{\Delta E}{2} = (E_{out}-E_{in})/2 = E_{out}/2 = k_e \frac{Q}{2R^2}$$ again in agreement with the result you found. (For more details on this I suggest Griffith's book on electrodynamics sec. 2.4 and 2.5)

Now to compute the energy stored in a given configuration it is enough to realize that, the energy stored in the electric field of a point charge is already infinite, since it somehow measures how much energy went into building the configuration so you would need to drag let's say some charge $Q$ and accumulate it into a point and Coulomb's law already tells us the force would be infinite.

So when you use the discrete version for computing the energy, one is not taking into consideration the energy needed to build the point charges themselves, reason why you demand $i\neq j$. While in the continuous version of the formula $\Phi$ is the total potential evaluated at the same point as the density, $\rho(\vec{r})\Phi(\vec{r})$, so you expect this quantity to already be infinite if not taken with care, because it is implicitly accounting for the energy of building up each "charge point". Again if you read sec. 2.5 from Griffith's book, you will find some more details.

Let us first compute the electric field on the surface of a spherical (infinitely thin) shell with a uniform charge distribution. We can use Gauss's law for this, $$\int_{S_r} \vec{E}\cdot d\vec{A} = \frac{Q_{in}}{\varepsilon_0},$$ from which using that the charge density is $\sigma = Q/4\pi R^2$ and taking the Gauss surface at $r=R$, we get (using the spherical symmetry): $$E = \frac{4\pi R^2\sigma}{4\pi r^2\varepsilon_0} \stackrel{r\longrightarrow R}{=} k_e\frac{Q}{R^2}$$ as you stated yourself. However notice that for all things to hold, the limit is taken from above, namely the Gaussian surface is taken from outside and reduced. On the other hand we also know the field inside will be zero, so there is a clear discontinuity in the electric field at $r=R$. So formally the field is not defined there, but we regularize the field by taking the average of the discontinuity jump. In this case, $$E(R):= \frac{\Delta E}{2} = (E_{out}+E_{in})/2 = E_{out}/2 = k_e \frac{Q}{2R^2}$$ again in agreement with the result you found. (For more details on this I suggest Griffith's book on electrodynamics sec. 2.4 and 2.5)

Now to compute the energy stored in a given configuration it is enough to realize that, the energy stored in the electric field of a point charge is already infinite, since it somehow measures how much energy went into building the configuration so you would need to drag let's say some charge $Q$ and accumulate it into a point and Coulomb's law already tells us the force would be infinite.

So when you use the discrete version for computing the energy, one is not taking into consideration the energy needed to build the point charges themselves, reason why you demand $i\neq j$. While in the continuous version of the formula $\Phi$ is the total potential evaluated at the same point as the density, $\rho(\vec{r})\Phi(\vec{r})$, so you expect this quantity to already be infinite if not taken with care, because it is implicitly accounting for the energy of building up each "charge point". Again if you read sec. 2.5 from Griffith's book, you will find some more details.