Baltic Way competition problems, 2019

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

It remains to show that all natural numbers smaller than $m$ have a representation in the form $\binom{a}{2} + \binom{b}{2} + c$ 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 $\binom{a}{2} + \binom{b}{2} + (c-1)$ or $\binom{a}{2} + \binom{b-1}{2} + (b-2)$, respectively. Hence, we can represent all integers between $\binom{a}{2}$ and $k$ in the desired form. Therefore and because we have a representation for $m-1$ already, it suffices to show that the numbers $\ell_s = \binom{1957-s}{2} - 1$ with $s = 0, 1, \dots, 1954$ can be represented. Now the claim follows by $\ell_0 = \binom{1956}{2} + \binom{63}{2} + 2$ and $\binom{1956-s}{2} < \ell_s \le \binom{1956-s}{2} + \binom{b}{2}$ with $b = \min\{63, 1956-s\}$ for $s = 1, 2, \dots, 1954$.