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 $(a,b)$.
