I just thought about the following question: Find all functions $g:\mathbb{N}\to\mathbb{Z}$ such that $m+n~|~g(m)+g(n)$ for all $m,n\in\mathbb{N}$.
Clearly every polynomial $g(X)\in\mathbb{Z}[X]$ in which $X$ only occurs to an odd power is a solution.
Taking $m=n$ gives $m~|~g(m)$ and hence, introducing $h(m)=m\cdot g(m)$, we can rewrite the condition as $m+n~|~n(h(m)-h(n))$ for all naturals $m$ and $n$. This in turn is equivalent to $m+n~|~h(tm)-h(tn)$ for all naturals $m,n,t$.
But now I am stuck. Do you have any ideas or solutions? Do you think/spot/know that there are more solutions than the mentioned polynomials?
First, note that the set of functions satisfying your condition has the property that it is closed under integer linear combinations. Give two such functions, $p,q,$ and integers $a,b$, the function $r(n)=ap(n)+bq(n)$ has the above property.
Let $p_k(x)=x(x^2-1)\dots(x^2-k^2)$. This is a polynomial with only odd degrees.
Then for any sequence of integers $a_k$, you can define:
$$p(n)=\sum_{k=0}^\infty a_kp_k(n)$$
This is well defined because for $k>n$, $p_k(n)=0$, so it is, locally, a finite sum for any $n$. (Left to the reader: show $p(n)$ has your property. It does.)
This means that there are uncountably many such functions.
This is related to an old question of mine: finding all $f:\mathbb N\to\mathbb Z$ with $m-n\mid f(m)-f(n)$. There, I answered my own question, but I think your question might be slightly different. Still, the above is "like" what I did in that older question, so possibly worth checking out.
In particular, if $g:\mathbb Z\to\mathbb Z$ is an odd function with $\forall m,n:m-n\mid f(m)-f(n)$, then $g$ restricted to the natural numbers is in your class. These functions are classified by a sequence of integers, $a_k$ using the functions:
$$\sum_{k=0}^\infty a_k\frac{\mathrm{lcm}\{1,\dots,2k+1\}}{(2k+1)!}p_k(x)$$
But your condition is really that such odd function satisfy $m-n\mid f(m)-f(n)$ only for $n<0$ and $m>0$. So that might allow for more functions.
But I think I can outline a proof much like my proof for my question. Namely, if we have the values $f(0),\dots,f(n)$, that determines the value of $f(n+1)$ modulo ${\mathrm{lcm}\{n+1,\dots,2n+1\}}$ since we know:
$$f(n+1)\equiv -f(d)\pmod{n+1+d}$$ for $d=0,\dots,n$. (The case $d=n+1$ follows from the case $d=0$.)
But $\mathrm{lcm}\{n+1,\dots,2n+1\} = \mathrm{lcm}\{1,\dots,2n+1\}$.
So we can inductively find $a_k$ so that:
$$f(n)=\sum_{k=0}^\infty a_k\frac{\mathrm{lcm}\{1,\dots,2k+1\}}{(2k+1)!}p_k(n)$$
This also means that your condition implies $\forall m,n.m-n\mid g(m)-g(n)$ too. I wondered if there was a direct proof of this, and I asked here and got a quick easy proof.
That easy proof lets you re-use my result directly when classifying your functions, since it means that your $g$ can be extended to a function $g_1:\mathbb Z\to\mathbb Z$ with $m-n\mid g_1(m)-g_1(n)$ and $g_1(-m)=-g_1(m)$. So that lets us use my classification result.