[Number Theory]
Problem: Compute the least positive integer $n$ such that the sum of the digits of $n$ is five times the sum of the digits of $(n+2019)$.

Solution: Let $S(n)$ denote the sum of the digits of $n$, so that solving the problem is equivalent to solving $S(n)=5 S(n+2019)$. Using the fact that $S(n) \equiv n(\bmod 9)$ for all $n$, it follows that

$$
\begin{aligned}
n & \equiv 5(n+2019) \equiv 5(n+3)(\bmod 9) \\
4 n & \equiv-15(\bmod 9) \\
n & \equiv 3(\bmod 9)
\end{aligned}
$$

Then $S(n+2019) \equiv 6(\bmod 9)$. In particular, $S(n+2019) \geq 6$ and $S(n) \geq 5 \cdot 6=30$. The latter inequality implies $n \geq 3999$, which then gives $n+2019 \geq 6018$. Thus if $n+2019$ were a four-digit number, then $S(n+2019) \geq 7$. Moreover, $S(n+2019)$ can only be 7, because otherwise, $S(n)=5 S(n+2019) \geq 40$, which is impossible (if $n$ has four digits, then $S(n)$ can be no greater than 36). So if $n+2019$ were a four-digit number, then $S(n+2019)=7$ and $S(n)=35$. But this would imply that the digits of $n$ are $8,9,9,9$ in some order, contradicting the assumption that $n+2019$ is a four-digit number. On the other hand, if $n+2019$ were a five-digit number such that $S(n+2019) \geq 6$, then the least such value of $n+2019$ is 10005 , and indeed, this works because it corresponds to $n=\mathbf{7 9 8 6}$, the least possible value of $n$.

Final answer: 7986