For each $k=1,2,3,4$, note that set of pairs $\left(a_{k}, b_{k}\right)$ with $0 \leq a_{k} \leq k$ maps bijectively to the set of nonnegative integers through the map $\left(a_{k}, b_{k}\right) \mapsto a_{k}+(k+1) b_{k}$, as $a_{k}$ is simply the remainder of $a_{k}+(k+1) b_{k}$ upon division by $k+1$. By letting $x_{k}=a_{k}+(k+1) b_{k}$, we see that the problem is equivalent to finding the number of quadruples of nonnegative integers $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ such that $x_{1}+x_{2}+x_{3}+x_{4}=19$. This is the same as finding the number of quadruples of positive integers $\left(x_{1}+1, x_{2}+1, x_{3}+1, x_{4}+1\right)$ such that $x_{1}+x_{2}+x_{3}+x_{4}=23$. By a standard "dots and bars" argument, we see that the answer is $\binom{22}{3}=1540$.