 Solution

Take an independent and identically distributed sequence with $X_i\sim \mathrm{Bernoulli}(r)$, independent of $N$. Conditional on $N=n$, the random sum ${\sum }_{i=1}^n{X}_i$ has binomial distribution with parameters $n,r$, including the empty sum at $n=0$. This construction therefore has exactly the prescribed joint law of $(N,Z)$ and represents $Z$ as the required random sum.

For a literal representation that retains an already given pair $(N,Z)$, enlarge the probability space by auxiliary randomness. Conditional on $N=n,Z=z$, choose uniformly a size-$z$ subset of $\{1,\dots ,n\}$, put ones at those indices and zeros at the others, and append independent Bernoulli variables of parameter $r$. Given $N=n$, a specified length-$n$ pattern with $z$ ones has probability $\left(\genfrac{}{}{0ex}{}{n}{z}\right)r^z(1-r{)}^{n-z}/(\genfrac{}{}{0ex}{}{n}{z})=r^z(1-r{)}^{n-z}$. Thus the entire sequence is independent and identically distributed and independent of $N$, while its first $N$ terms sum to the original $Z$.