= Solution
A <highest-weight vector> is a nonzero <weight vector> $v$ such that $\mathfrak g_\alpha v=0$ for every <positive root> $\alpha$. Its <weight of a representation> is a highest weight. For an irreducible finite-dimensional <Lie algebra representation>, this is the unique <highest weight of a representation>, and $v$ generates the module; a reducible module can have several highest weights of different summands.
Here is an existence argument for every nonzero finite-dimensional module $V$. For each <simple root>, its root operators and <coroot> form an <sl2 subalgebra associated with a root>. The <classification of finite-dimensional sl2 representations> shows that this coroot acts diagonalizably with integral <eigenvalues>. These commuting coroots span the <Cartan subalgebra>, so simultaneous diagonalization gives a finite <weight-space decomposition>
$$
V=\bigoplus_\lambda V_\lambda.
$$
Choose a real linear functional $\ell$ on the <weight lattice> span that is positive on every <simple root>, hence on every <positive root>, and choose a weight $\lambda$ maximizing $\ell$ among the finitely many weights of $V$. If $v\in V_\lambda$ and $X_\alpha\in\mathfrak g_\alpha$, then
$$
H(X_\alpha v)=X_\alpha Hv+[H,X_\alpha]v=(\lambda(H)+\alpha(H))X_\alpha v,
$$
so $X_\alpha v$ belongs to $V_{\lambda+\alpha}$ if nonzero. For positive $\alpha$, that would contradict maximality because $\ell(\lambda+\alpha)>\ell(\lambda)$. Thus every nonzero $v\in V_\lambda$ is a <highest-weight vector>. \b[Every nonzero finite-dimensional module therefore has a highest weight.] The zero module has no nonzero vector or weight, so it is the vacuous exception to the literal unrestricted wording.
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