Sharpness of Cramer-Wold: A pair $(X,Y)$ where a finite set of projections are normal but $(X,Y)$ not joint normal In the case of normality, the Cramer-Wold Theorem states:

Let $(X,Y)$ be a bivariate random variable.  If for all $a,b \in \mathbb{R}$, $aX + bY$ is Gaussian, then $(X,Y)$ is jointly Gaussian.

I'm aware that Cramer-Wold can be improved: you only need to check a countably infinite collection of pairs $(a,b)$.  My question is: 

Let $(a_i,b_i)_{i = 1}^N$ be a finite collection of pairs of real numbers.  Can you construct a bivariate random variable $(X,Y)$ so that $a_i X + b_i Y$ is Gaussian for each $i$, but $(X,Y)$ is not jointly Gaussian?

I'm struggling to come up with a robust way of making counterexamples for this problem; I know the classic example when we want $X$ and $Y$ to be normal but not jointly normal, but am stuck here.  Any help would be appreciated.
 A: Notation: Let $\alpha_i=(a_i,b_i)\in\mathbb R^2$, let $\tau_i=(-b_i, a_i)\in \mathbb R^2$.  Note that $\langle\alpha_i,\tau_i\rangle=0$ for all $1\le i\le N$.  What is sought is a non-gaussian random vector $Z$ with distribution $\gamma^*$ such that all $\langle\alpha_i,Z\rangle$ are  gaussian.  I will construct an absolutely continuous compactly supported signed measure $\nu$ with bounded density on $\mathbb R^2$ all of whose margins in the directions $\alpha_i$ vanish.  Then some small multiple of $\nu$ added to a nondegenerate bivariate gaussian measure $\gamma$ (such as the one with joint density $\phi(x)\phi(y)$) will do the trick.
Consider the discrete signed measure $\mu$ which is the convolution of the signed measures $(\delta_{\tau_i}-\delta_{-\tau_i})/2$, for  $1\le i\le N$. Its Fourier transform is $\psi(u)=\int_{\mathbb R^2} \exp(i\langle u,x\rangle)\mu(dx) = \prod_{i=1}^N \sin( \langle u,\tau_i\rangle).$
All sections through $\psi$ in directions $\alpha_i$ vanish: $\forall i, t\mapsto \psi(t\alpha_i)$ is the constant zero function.  That is, the marginals of $\mu$ in the directions $\alpha_i$ all vanish.  (In effect the 
 $2^N$ signed point masses in the support of $\mu$ hide behind each other with cancelling signs when viewed in the directions $\alpha_i$.)  Now convolve $\mu$ with (say) the indicator function of a small disk, to obtain $\nu$. As indicated above, the density of $\nu$ is compactly supported and bounded, so there exists a sufficiently small $\epsilon>0$ such that $\gamma^*=\gamma+\epsilon\nu$ is a probability measure (that is, non-negative).   A graph of the density function of $\gamma^*$ looks like an exhibit at Legoland: it looks like that of $\gamma$ with added disk-shaped knobs and craters.
(As a note: what one convolves $\mu$ with to produce $\nu$ does not much matter, and the recipe given here is perhaps overly prescriptive.)
