They may be dependent. Put $X_1 = X_2 = 0$, $X_3$ - Bernoulli random Variable.
There's a notion "n independent random variables", which may be found in every book on probability.
Addition: if $Y_1$ and $Y_2$ are independent then the characteristic function of $(X_1 - X_3, X_2 - X_3)$ is
$$Ee^{i \bigl( a(X_1-X_3) + b(X_2-X_3) \bigr) } = Ee^{i a(X_1-X_3)} \cdot Ee^{i b(X_2-X_3)} \ \ \ (\star)$$ As $X_1, X_2, X_3$ are independent left-hand side of ($\star$) is $$Ee^{i \bigl( a X_1 + b X_2 -(a+b) X_3 \bigr) }=Ee^{i a X_1} Ee^{i b X_2} Ee^{i ( -(a+b) X_3)}$$ and right-hand side of $(\star)$ is $$Ee^{i a X_1 } Ee^{i (-a)X_3} Ee^{i b X_2} Ee^{i (-b)X_3}.$$ Thus $$Ee^{i a X_1} Ee^{i b X_2} Ee^{i ( -(a+b) X_3)} = Ee^{i a X_1 } Ee^{i (-a)X_3} Ee^{i b X_2} Ee^{i (-b)X_3}.$$ Suppose that $Ee^{i a X_1} = 0$ only for $a$ from a set of measure $0$. Also suppose that $Ee^{i b X_2} = 0$ only for $b$ from a set of measure $0$. These assumptions holds true for all popular distributions. We have $Ee^{i ( -(a+b) X_3)} = Ee^{i (-a)X_3} Ee^{i (-b)X_3}$ for almost all $a$ and $b$. Characteristic functions are continious, hence $Ee^{i ( -(a+b) X_3)} = Ee^{i (-a)X_3} Ee^{i (-b)X_3}$ for all $a$ and $b$. Put $t = -a, s = -b$. Thus $$Ee^{i (t+s) X_3} = Ee^{i t X_3} Ee^{i s X_3}$$ for all $t$ and $s$. We got that the characteristic function of a vector $(X_3, X_3)$ is equal to the product of characteristic functions of it's components. Hence $X_3$ and $X_3$ are independent and hence $X_3 = const$.
If $X_3 = const$ then $Y_1$ and $Y_2$ are obviously independent.
So the condition that you need is the condition $X_n = const$.
Remark: there are characteristic functions which are equal to $0$ on the set of positive measure, e.g. $(1-|t|)I_{|t| < 1}$. It follows from Polya theorem about characteristic functions.
