Is joint normality a necessary condition for the sum of normal random variables to be normal? In comments following this answer of mine to a related question, Users ssdecontrol and Glen_b asked
whether joint normality of $X$ and $Y$ is necessary for asserting the
normality of the sum $X+Y$? That joint normality is sufficient is,
of course, well-known. This supplemental question was not addressed
there, and is perhaps worth considering in its own right.
Since joint normality implies marginal normality, I ask

Do there exist normal random variables $X$ and $Y$ such that
  $X+Y$ is a normal random variable, but $X$ and $Y$ are not
  jointly normal random variables?

If $X$ and $Y$ are not required to have normal distributions,
then it is easy to find such normal random variables. One example
can be found in my previous answer (link is given above).
I believe that the answer to the highlighted question above
is Yes, and have posted (what I think is) an example as an answer
to this question.
 A: Consider jointly continuous random variables $U, V, W$  with joint density function 
\begin{align}
f_{U,V,W}(u,v,w) = \begin{cases} 2\phi(u)\phi(v)\phi(w)
& ~~~~\text{if}~ u \geq 0, v\geq 0, w \geq 0,\\
& \text{or if}~ u < 0, v < 0, w \geq 0,\\
& \text{or if}~ u < 0, v\geq 0, w < 0,\\
& \text{or if}~ u \geq 0, v< 0, w < 0,\\
& \\
0 & \text{otherwise}
\end{cases}\tag{1}
\end{align}
where $\phi(\cdot)$ denotes the standard normal density function.
It is clear that $U, V$, and $W$ are dependent
random variables. It is also clear that they are not
jointly normal random variables.
However, all three pairs $(U,V), (U,W), (V,W)$
are pairwise independent random variables: in fact,
independent standard normal random variables (and thus
pairwise jointly normal random variables).
In short,
$U,V,W$ are an example of pairwise independent but not
mutually independent normal random variables.
See this answer of mine
for more details.
Notice that the pairwise independence gives us that
$U+V, U+W$, and $V-W$ all are zero-mean normal random
variables with variance $2$. Now, let us define
$$X = U+W, ~Y = V-W \tag{2}$$ and note that $X+Y = U+V$
is also a zero-mean normal random variable with variance $2$.
Also, $\operatorname{cov}(X,Y) = -\operatorname{var}(W) = -1$,
and so $X$ and $Y$ are dependent and correlated random variables.

$X$ and $Y$ are (correlated) normal random variables that are not
  jointly normal but have the property that their sum $X+Y$ is
  a normal random variable.

Put another way, joint normality is a sufficient condition for 
asserting the normality of a sum of normal random variables,
but it is not a necessary condition.
Proof that $X$ and $Y$ are not jointly normal
Since the transformation $(U,V,W) \to (U+W, V-W, W) = (X,Y,W)$ is
linear, it is easy to get that 
$f_{X,Y,W}(x, y, w) = f_{U,V,W}(x-w,y+w,w)$.
Therefore we have that
$$f_{X,Y}(x,y) = \int_{-\infty}^\infty f_{X,Y,W}(x,y,w)\,\mathrm dw
= \int_{-\infty}^\infty f_{U,V,W}(x-w,y+w,w)\,\mathrm dw$$
But $f_{U,V, W}$ has the property that its value is nonzero only
when exactly one or all three of its arguments are nonnegative.
Now suppose that $x, y > 0$. Then, $f_{U,V,W}(x-w,y+w,w)$ has
value $2\phi(x-w)\phi(y+w)\phi(w)$ for 
$w \in (-\infty,-y) \cup (0,x)$ and is $0$ otherwise. So,
for $x, y > 0$,
$$f_{X,Y}(x,y) 
= \int_{-\infty}^{-y} 2\phi(x-w)\phi(y+w)\phi(w)\,\mathrm dw
+ \int_0^x 2\phi(x-w)\phi(y+w)\phi(w)\,\mathrm dw.\tag{3}$$
Now, 
\begin{align}
(x-w)^2 + (y+w)^2 + w^2 &= 3w^2 -2w(x-y) + x^2 + y^2\\
&= \frac{w^2 - 2w\left(\frac{x-y}{3}\right) 
+ \left(\frac{x-y}{3}\right)^2}{1/3}
-\frac 13(x-y)^2 + x^2 + y^2
\end{align}
and so by expanding out $2\phi(x-w)\phi(y+w)\phi(w)$ and
doing some re-arranging of the integrands in $(3)$, we
can write
$$f_{X,Y}(x,y) = g(x,y)\big[P\{T \leq -y\} + P\{0 < T \leq x\}\big]
\tag{4}$$ 
where $T$ is a normal random variable with mean $\frac{x-y}{3}$
and variance $\frac 13$. Both terms inside the square brackets
involve the standard normal CDF $\Phi(\cdot)$ with arguments
that are (different) functions of both $x$ and $y$. Thus, $f_{X,Y}$ is
not a bivariate normal density even though both $X$ and $Y$
are normal random variables, and their sum is a normal random variable.

Comment: Joint normality of $X$ and $Y$ suffices for normality
of $X+Y$ but it also implies much much more: $aX+bY$ is normal for
all choices of $(a,b)$. Here, we need $aX+bY$ to be normal
for only three choices of $(a,b)$, viz., $(1,0), (0,1), (1,1)$
where the first two enforce the oft-ignored condition (see e.g. the
answer by $Y.H.$) that
the (marginal) densities of $X$ and $Y$ must be normal densities,
and the third says that the sum must also have a normal density.
Thus, we can have normal random variables that are not
jointly normal but whose sum is normal because we don't care
what happens for other choices of $(a,b)$.
A: Let $U,V$ be iid $N(0,1)$. 
Now transform $(U,V) \to (X,Y)$ as follows:
In the first quadrant (i.e. $U>0,V>0$) let $X=\max(U,V)$ and $Y = \min(U,V)$.
For the other quadrants, rotate this mapping about the origin.
The resulting bivariate distribution looks like (seen from above):
$\hspace{1.5 cm}$
-- the purple represents regions with doubled probability and the white regions are ones with no probability. The black circles are contours of constant density (everywhere on the circle for $(U,V)$, but within each colored region for $(X,Y)$).


*

*By symmetry both $X$ and $Y$ are standard normal (looking down a vertical line or along a horizontal line there's a purple point for every white one which we can regard as being flipped across the axis the horizontal or vertical line crosses)

*but $(X,Y)$ are clearly not bivariate normal, and 

*$X+Y = U+V$ which is $\sim N(0,2)$ (equivalently, look along lines of constant $X+Y$ and see that we have symmetry similar to that we discussed in 1., but this time about the $Y=X$ line)
