If we consider the weights of a complex irreducible representation $V$ of a complex semisimple Lie algebra $\mathfrak{g}$, with respect to a chosen Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$, so that the weights of $V$ are elements of $\mathfrak{h}^*$, each with some multiplicity (which could be $1$ or greater), then does there always exist some complex representation $W$ of $SL(N)$, for some positive integer $N$, and a complex linear map $f$ from $\mathfrak{h}$ to a Cartan subalgebra $\mathfrak{k}$ of $SL(N)$, such that the pull-backs $f^*(\mu)$ (with $\mu$ being a weight of $W$) of the weights of $W$ are precisely the weights of $V$, taking into account multiplicity?
Edit: on second thought, partly due to @DavidESpeyer's question below, I realize now that I am interested in the following category, where the objects are pairs $(V, \Phi)$, where $V$ is a finite-dimensional real vector space and $\Phi$ is a finite multiset consisting of elements of $V^*$. A homomorphism from $(V_1, \Phi_1)$ to $(V_2, \Phi_2)$ consists of a linear map $f: V_1 \to V_2$ such that $f^*(\Phi_2) \subseteq \Phi_1$ (in the sense that $f^*(\Phi_2)$ is a sub-multiset of $\Phi_1$).
There is a distinguished family of such objects, consisting of $(\mathbb{R}^n, (x_i))$, where the $x_i$, for $i = 1, \ldots, n$, are the (standard) coordinate functions on $\mathbb{R}^n$, each with multiplicity $1$.
Question: given an object $(V, \Phi)$, does there always exist some positive integer $n$ and a homomorphism from $(V, \Phi)$ to $(\mathbb{R}^n, (x_i))$? I believe so. I think one can label the elements of $\Phi$, taking into account multiplicity, from $1$ to $n$, say, where $n$ is the order of $\Phi$. We then map $x_i$ to the element of $\Phi$ labeled by $i$, say $\alpha_i$, for $i = 1, \ldots, n$. This defines a linear map from $(\mathbb{R}^n)^*$ to $V^*$. Its dual, is then a linear map from $(V^*)^*$ to $((\mathbb{R}^n)^*)^*$. But the double dual of a finite-dimensional vector space is canonically isomorphic to itself (using the evaluation map), so we have defined a homomorphism from $(V, \Phi)$ to $(\mathbb{R}^n, (x_i))$. Ok, this is what I was wondering about. I guess I answered it (unless I made a silly mistake somewhere). Also see @DavidESpeyer's answer below, to the question I had asked before I wrote down this edit!
$\def\fh{\mathfrak{h}}\def\fg{\mathfrak{g}}\def\sl{\mathfrak{sl}}$I suspect this isn't what you mean to ask but, if you really did, the answer is yes.
Let $\lambda_1$, $\lambda_2$, ..., $\lambda_N$ be any elements of $\fh^{\ast}$ with $\sum \lambda_i = 0$. Map $\fh \to \sl(N)$ by $$h \mapsto \text{diag}(\lambda_1(h), \lambda_2(h), \ldots, \lambda_N(h)).$$ (I need $\sum \lambda_i=0$ so that the map lands in $\sl(N)$, not just $\mathfrak{gl}(N)$.) Let $W$ be the standard $N$-dimensional representation of $\sl(N)$. Then the weights of $\sl(N)$, restricted to $\fh$, are the $\lambda_i$.
It remains to check that, if $\fg$ is a semisimple Lie algebra, $V$ a representation of $\fg$, and $\fh$ a Cartan of $\fg$, then the weights of $\fh$ acting on $V$ sum to $0$. The sum of those weights gives the weight of $\fh$ acting on the one dimensional vector space $\bigwedge^{\dim V} V$. Since $\fg$ is semisimple, it has no abelian quotient, and the action of $\fg$ on $\bigwedge^{\dim V} V$ is thus trivial.