Convolution that results in an all-zero sequence
Expanding on my comment when this question was closed, suppose that
$$x[n]=e^{j2\pi n/3},~\forall n,\quad y[n]=\begin{cases}e^{j4\pi n/3}&amp; n=0,1,2,\\0, &amp;\text{otherwise.}\end{cases}$$

Then,
$\displaystyle \sum_{n=-\infty}^\infty x[k−n]y[n]= \sum_{n=0}^2 e^{j2\pi (k−n)/3}e^{j4\pi n/3}
= e^{j2\pi k/3}\sum_{n=0}^2 e^{j2\pi n/3}=0.$

More generally,   $\displaystyle x[n]=e^{j2\pi n/N},~\forall n,\quad y[n]=\begin{cases}e^{j2\pi (N-n)/N}&amp; n=0,1,2,\ldots, N-1\\0, &amp;\text{otherwise,}\end{cases}$
works as a solution for sequences taking on $N$ values.
Of course, nitpickers might argue that y takes on $N+1$ values, not $N$
as required by the OP, but in the case $N = 3$, there is an easy workaround.
Define 
$$z[n] = \begin{cases}1, &amp; n = 0,\\
-e^{j2\pi/3}, &amp; n = 1,\\0, &amp;\text{otherwise,}\end{cases}$$
which is clearly a three-valued sequence, and note that
$$\displaystyle \sum_{n=-\infty}^\infty x[k−n]z[n]= 
\sum_{n=0}^1 e^{j2\pi (k-n)/3}z(n) = e^{j2\pi k/3}\cdot 1
+ e^{j2\pi (k-1)/3}\cdot (-e^{j2\pi/3})
= e^{j2\pi k/3} - e^{j2\pi k/3} = 0.$$ The real pezzonovante nitpickers
who wish to complain that the three values that $z$ takes on are
not the same three values that $x$ takes on are invited to work out
a more satisfactory solution for themselves.

It is tempting to extend $y$ periodically
to be nonzero for all $n$ so as to totally avoid the nitpick, but unfortunately,
the infinite sum $\sum_{n=-\infty}^\infty e^{j2\pi n/3}$ that is the
result of the convolution is not absolutely
convergent.  Of course, since both sequences are of period $3$, their
 convolution could also be defined as a periodic convolution,
which is the sum of just $3$ terms, and we are back in business.
More generally, note that the periodic
convolution of any two rows of the $N\times N$ DFT matrix is $0$,
and so we can extend this notion to sequences that take on $N$
distinct values for any $N \geq 3$.  For another viewpoint on this
approach, see the answer by @Hilmar.