Are the goods in additively separable utility functions normal goods? Inspired by this answer.

To make it a bit more precise, by normal good I mean demand is (not necessarily strictly) increasing in income, and by additively separable utility function I mean that a monotone transformation exists for which
$$
U(x_1,x_2,\dots) = f_1(x_1) + f_2(x_2) + \dots 
$$
(Also assume $U$ is increasing in all variables.) 
Given a linear budget constraint and a utility maximizing consumer, do these goods $x_i$ exhibit normal behavior for all income levels?
Yes if you assume that the sub-utility functions are concave. Notice that this is a standard assumption as otherwise, the utility function $u = \sum_i f_i$ is not guaranteed to be concave (nor quasi-concave).
Let denote by $u_i = \dfrac{\partial u}{\partial x_i}$ and by $u_{i,j} = \dfrac{\partial^2 u}{\partial x_i \partial x_j}$. By additivity $u_{i,j} = 0$ if $i \ne j$.
The first order conditions for the utility maximisation problem give:
$$
\begin{align*}
u_i = \lambda p_i, \tag{1}\\
\sum_i p_i x_i = m \tag{2}
\end{align*}
$$
Differentiating both with respect to income $m$ (and using $u_{i,j} = 0$) gives:
$$
\begin{align*}
&amp;u_{ii} \frac{\partial x_i}{\partial m} = p_i \frac{\partial \lambda}{\partial m} \tag{3}\\
&amp;\sum_i p_i \frac{\partial x_i}{\partial m} = 1 \tag{4}
\end{align*}
$$
Substitute $(3$) into $(4)$:
$$
\begin{align*}
&amp;\sum_i (p_i)^2 \frac{\partial \lambda}{\partial m} \frac{1}{u_{ii}} = 1,\\
\to &amp;\frac{\partial \lambda}{\partial m} = \frac{1}{\sum_i \frac{(p_i)^2}{u_{ii}}} \tag{5}
\end{align*}
$$
Notice that $u_{ii} &lt; 0$ by concavity of the sub-utility functions. As such, $\frac{\partial \lambda}{\partial m} &lt; 0$ and also, by $(3)$:
$$
\frac{\partial x_i}{\partial m} = p_i \frac{1}{u_{ii}} \frac{\partial \lambda}{\partial m} &gt; 0 
$$
as the right hand side is the product of two negative numbers.
This shows that $x_i$ is a normal good. By symmetry, this holds for all goods.
An alternative quicker way to notice this is to see that

by $(3)$ the sign of $\dfrac{\partial x_i}{\partial m}$ will be the reverse of the sign of $\dfrac{\partial \lambda}{\partial m}$.
This has to be true for all goods, which means that either all goods are normal, or all goods are inferior.
From $(4)$, it follows that at least one good should be normal (as otherwise the sum cannot be equal to 1 which is greater than zero).
Conclude that all goods have to be normal.