Baltic Way 2019

The number is $m=\left(\genfrac{}{}{0ex}{}{1957}{2}\right)+\left(\genfrac{}{}{0ex}{}{63}{2}\right)+1=1,915,900$. Assume that $m$ has a representation as above. Then $a\le 1957$ as $\left(\genfrac{}{}{0ex}{}{1958}{2}\right)>m$. On the other hand, by $\left(\genfrac{}{}{0ex}{}{a}{2}\right)+\left(\genfrac{}{}{0ex}{}{b}{2}\right)+c\le \left(\genfrac{}{}{0ex}{}{a+1}{2}\right)+\left(\genfrac{}{}{0ex}{}{b-1}{2}\right)+(b-1)$ it follows that the largest number that can be represented as above with $a\le 1957$ is $\left(\genfrac{}{}{0ex}{}{1957}{2}\right)+\left(\genfrac{}{}{0ex}{}{62}{2}\right)+62=m-1$, a contradiction.

It remains to show that all natural numbers smaller than $m$ have a representation in the form $\left(\genfrac{}{}{0ex}{}{a}{2}\right)+\left(\genfrac{}{}{0ex}{}{b}{2}\right)+c$ with $a\ge b\ge c$ and $a+b\le 2019$. If some number $k$ has such a representation and $c>0$ or $c=0,b\ge 2$, then $k-1$ can be represented as $\left(\genfrac{}{}{0ex}{}{a}{2}\right)+\left(\genfrac{}{}{0ex}{}{b}{2}\right)+(c-1)$ or $\left(\genfrac{}{}{0ex}{}{a}{2}\right)+\left(\genfrac{}{}{0ex}{}{b-1}{2}\right)+(b-2)$, respectively. Hence, we can represent all integers between $\left(\genfrac{}{}{0ex}{}{a}{2}\right)$ and $k$ in the desired form. Therefore and because we have a representation for $m-1$ already, it suffices to show that the numbers ${\ell }_s=\left(\genfrac{}{}{0ex}{}{1957-s}{2}\right)-1$ with $s=0,1,\dots ,1954$ can be represented. Now the claim follows by ${\ell }_0=\left(\genfrac{}{}{0ex}{}{1956}{2}\right)+\left(\genfrac{}{}{0ex}{}{63}{2}\right)+2$ and $\left(\genfrac{}{}{0ex}{}{1956-s}{2}\right)<{\ell }_s\le \left(\genfrac{}{}{0ex}{}{1956-s}{2}\right)+\left(\genfrac{}{}{0ex}{}{b}{2}\right)$ with $b=\min \{63,1956-s\}$ for $s=1,2,\dots ,1954$.