This is maybe a little basic for MathOverflow, but I'm hoping it will get some interesting answers.
Let $\unrhd$ be the dominance order on partitions of $n \in \mathbb{N}$.
For partitions $\lambda$ and $\mu$ of $n$, the Kostka Number $K_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $\mu$.
If $t$ is such a tableau then the $\mu_1+\cdots +\mu_r$ entries $k$ of $t$ such that $k \le r$ all lie in the first $r$ rows of $t$. Hence $\lambda_1 + \cdots + \lambda_r \ge \mu_1 + \cdots + \mu_r$ for each $r$ and so $\lambda \unrhd \mu$.
Is there a short combinatorial proof of the converse: if $\lambda \unrhd \mu$ then $K_{\lambda\mu} > 0$?
A constructive proof, maybe using the characterization of neighbours in the dominance order by single box moves on Young diagrams, would be especially welcome.
Edit. Using some representation theory it's possible to make this strategy work. The following is the symmetric functions version of Theorem 2.2.20 in the textbook by James and Kerber. Obviously $K_{\lambda\lambda} = 1$. Let $\mu$ and $\mu^\star$ be neighbours
in the dominance order, with $\mu \rhd \mu^\star$, so $\mu^\star_i=\mu_i-1$, $\mu^\star_j = \mu_j+1$ for some $i < j$, and $\mu^\star_k = \mu_k$ if $k\not=i,j$. Since $h_{(a,b)} = h_{(a+1,b-1)} + s_{(a,b)}$ whenever $a \ge b$, we have
$$h_{\mu^\star} = \bigl(\prod_{k\not=i,j} h_{\mu_k} \bigr) h_{\mu_i-1}h_{\mu_j+1}=
\bigl(\prod_{k\not=i,j}h_{\mu_k}\bigr) \bigl( h_{\mu_i}h_{\mu_j} + s_{(\mu_i-1,\mu_j+1)} \bigr) $$
and so if $f = \prod_{k\not=i,j}h_{\mu_k}$ then
$$K_{\lambda\mu^\star} = \langle s_\lambda, h_{\mu^\star} \rangle = \langle s_\lambda, h_\mu \rangle + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle = K_{\lambda\mu} + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle \ge K_{\lambda\mu}.$$
Best Answer
I think the following is a simple combinatorial argument which constructs the most dominant semistandard $\lambda$-tableau of content $\mu$ whenever $\lambda\trianglerighteq\mu$. (n.b. I haven't followed the reference given in Richard Stanley's comment, so I don't know whether I'm duplicating what's done there.)
In a nutshell, the idea is "put the largest numbers as low as possible". So let $l$ be the length of $\mu$, and start the tableau by putting the $\mu_l$ $l$s in the bottom boxes of columns as far to the left as possible subject to the condition that if any box has an $l$ and there is a box directly to the right, then that box must also have an $l$ in it. (Another way of saying this is: put $l$s at the bottom of the first $\mu_l$ columns, and then slide these $l$s to the ends of their rows.) If we let $j$ be maximal such that $\lambda_j\geq\mu_l$, this means that we are putting $\lambda_x-\lambda_{x+1}$ $l$s at the end of row $x$ for each $x>j$, and $\mu_l-\lambda_{j+1}$ $l$s at the end of row $j$.
To fill in the rest of the tableau, we work recursively. Let $\hat\lambda$ denote the partition whose Young diagram comprises the boxes that are still empty, and let $\hat\mu$ be the partition $(\mu_1,\dots,\mu_{l-1})$. Then as long as $\hat\lambda\trianglerighteq\hat\mu$, we can fill in the rest of the tableau with a semistandard $\hat\lambda$-tableau of content $\hat\mu$, which gives us what we need.
So we need to show that $\hat\lambda_1+\dots+\hat\lambda_x\geq\hat\mu_1+\dots+\hat\mu_x$ for every $x$. For $x<j$ or $x\geq l$ this is immediate from the fact that $\lambda\trianglerighteq\mu$, so take $j\leq x<l$. Observe that $$\lambda_1+\dots+\lambda_x=n-(\lambda_{x+1}+\dots+\lambda_l)\geqslant n-(l-x)\lambda_{x+1}$$while $$\mu_1+\dots+\mu_x=n-(\mu_{x+1}+\dots+\mu_l)\leqslant n-(l-x)\mu_l.$$ So $$(\hat\lambda_1+\dots+\hat\lambda_x)-(\hat\mu_1+\dots+\hat\mu_x)=(\lambda_1+\dots+\lambda_{x+1}-\mu_l)-(\mu_1+\dots+\mu_x)\geqslant(l-x-1)(\mu_l-\lambda_{x+1})\geqslant0$$ as required.