Method I: In such questions, we focus on comparing the numbers, so we avoid calculating their exact values.

From the equation, $p-2016=q+2017$, we can rewrite this as: $p=q+2017+2016$,which implies $p>q$.

From the equation $q+2017=r-2018$, we get: $r=q+2017+2018$, which implies $r>q$. From the equation $r-2018=s+2019$, we get: $r=s+2019+2018$, which implies $r>s$. From the equation $p-2016=r-2018$, we have: $r=p-2016+2018=p+2$, which implies $r>p$.

Since we know that $r>p$, $r>q$, and $r>s$, we can conclude that $r$ is the largest number.

Method II: Let $s=0$, so we get: $p-2016=q+2017=r-2018=2019$.

Now, solving each equation: $p-2016=2019$ implies $p=2019+2016=4035$, $q+2017=2019$ implies $q=2019-2017=2$, $r-2018=2019$ implies $r=2019+2018=4037$.

Thus, $r$ is the largest number.