We will prove a slightly stronger statement. Indeed, we'll show that there are infinitely many pairs $(m, n) \in \mathbb{N}^2$ satisfying the conditions above, as well as $m \equiv 99 \pmod{1000}$ and $n = m+9.$ Observe then that "we get the first condition for free," i.e. $n + s(2n) = m + s(2m)$ is implied by the constraints we have just added. So let's just show that for any $k \in \mathbb{Z}$, there are infinitely many pairs of positive integers $(m, n)$ so that $n = m+9$, $m \equiv 99 \pmod{1000},$ and $kn + s(n^2) = km + s(m^2)$. Call $k$ tasty if there is at least one such pair $(m, n),$ and tangy if there are infinitely many such pairs $(m, n).$ We seek to show that all positive integers are tangy. Lemma 1. There are arbitrarily small tasty integers. Proof. Consider $m = \frac{10^{2t+1} + 98198}{198}$ and $n = m+9$ for sufficiently large $t$. The corresponding $k$ can be as large as needed. $\blacksquare$ Lemma 2. If $k$ is tasty, then $k+1$ is tangy. Proof. Suppose that $k \in \mathbb{Z}$ is tasty. Let $(m, n)$ be a pair so that $n = m+9, m \equiv 99 \pmod{1000},$ and $kn + s(n^2) = km + s(m^2).$ Then consider $(10^{N} + m, 10^{N} + n)$ for sufficiently large $N$. These pairs imply that $k+1$ is tangy. $\blacksquare$ By the previous two lemmas, all integers are tangy and so we're done. $\square$

What does this s(n) mean exactly?

Sum of digits of n (Iin base 10)