Given a set, $X$, and a function $f: X \to \mathbb{N}$ (or indeed any totally ordered set), you can define a strict partial order where $x_0 < x_1$ iff $f(x_0) < f(x_1)$.
If $f$ is injective, this will be a total order. If not, it won't (because two elements in the pre-image of a number will be incomparable).
This is more restrictive than an arbitrary strict partial order. For example, take the strict partial order on $\{a, b, c, d\}$ generated by $a < b$, $a < c$ and $c < d$. If there were an $f: \{a,b,c,d\} \to \mathbb{N}$ that generated the order then we would have $f(b) = f(c)$ (since $b$ and $c$ are not comparable) and also that $f(b) = f(d)$ (since $b$ and $d$ are also not comparable). But then $f(c) = f(d)$ but I want $c < d$.
Is there a name for such an object and has anyone studied what properties it must have?
It’s a strict weak ordering. To see that every strict weak ordering can be obtained in this way, let $<$ be a strict weak order on a set $X$. For $x,y\in X$ let $x\sim y$ if and only if $x\not<y$ and $y\not<x$; the relation $\sim$ is transitive by hypothesis, and it is clearly reflexive and symmetric, so it’s an equivalence relation. Let $Y=X/{\sim}$, and let $f:X\to Y$ be the canonical quotient map.
Define a relation $\preceq$ on $Y$ by $f(x)\preceq f(y)$ if and only if $x<y$ or $x\sim y$; it’s straightforward to verify that $\preceq$ is a well-defined linear order and that $f(x)\prec f(y)$ if and only if $x<y$.