Tricky Absolute Value!

Firstly, note that the right-hand side of the given equation is even for all $n$ . This means that the left-hand side is also even, and so $2^m$ is odd; it follows that any solution must have $\boxed{m=0}$ , so the left-hand side must be $2020$ .

Because of the absolute value function on the right-hand side, we'll consider the cases $n \le 2020$ and $n>2020$ separately.

Case $n \le 2020$ :

The equation becomes $2020=(2020-n)+n-2020=0$ , which is clearly absurd. So there are no solutions in this case.

Case $n > 2020$ :

The equation is now $2020=(n-2020)+n-2020=2n-4040$ , which has the unique solution $\boxed{n=3030}$ . The only possible solution pair is $(m,n)=(0,3030)$ , giving the answer $\boxed{3030}$ .

For $n\leq 2020$ we have $2^{m} + 2019 = 0$ which has no solutions. For $n > 2020$ we have $2n - 2^{m} = 6059 \implies n = \frac{2^{m} + 6059}{2}$ $n$ can only be an integer when $2^{m}$ is odd, this only happens when $m=0$ and $n=3030$ .

So we’re left with $m+n = 0 + 3030 = \boxed{3030}$

First, it's clearly that RHS is even for all natural number of $n$ . Hence, LHS must be even too. Then, assume that $m \ge 1$ . From here, we get that $2^{m}+2019$ is odd, contradiction. Hence, $m=0$ , so we get $2020=|n-2020|+n-2020$ . Now, assume that $|n-2020|=2020-n$ . From here, we get that $2020=0$ , contradiction. Hence, $|n-2020|=n-2020$ , so we get $2020=2n-4040 \implies n=3030 \implies m+n=0+3030=3030$ .