They do. Simply choose, say, $X_n$ a Bernoulli random variable taking values $n$ or $-n$ with probability $\frac 12$ for each. Notice that $Var(X_n) = n^2$. Define $Y_n = - X_n$. You have $Var(X_n + Y_n) = 0$ and, to answer your last interrogation, $Cov(X_n,Y_n) = - Var(X_n) = -n^2$.

They do. Simply choose, say, $X_n$ a Bernoulli random variable taking values $n$ or $-n$ with probability $\frac 12$ for each. Notice that $Var(X_n) = n^2$. Define $Y_n = - X_n$. You have $Var(X_n + Y_n) = 0$ and, to answer your last interrogation, $Cov(X_n,Y_n) = - Var(X_n) = -n^2$.

Edit : Let now $Z$ be some random variable, as nice as you like (say, another symmetric Bernoulli, but on $\{-1,1\}$ this time) and independant from all the $X_n$. Choose $Y_n = - X_n + Z$. Then $Var(X_n + Y_n) \equiv 1$ and $Var(X_n) \sim Var(Y_n) \sim n^2$.