jmc

Let the two integers be $x$ and $246-x$. The product which needs to be maximized is $(x)(246-x)=246x-x^2$. Now we complete the square: $$\begin{array}{rl}-(x^2-246x)& =-(x^2-246x+15129-15129)\\ & =-(x-123{)}^2+15129\end{array}$$Since the square of a real number is always non-negative, $-(x-123{)}^2\le 0$. Therefore, the expression is maximized when $x-123=0$, so $x=123$. Therefore, the greatest product obtainable is $-(123-123{)}^2+15129=\boxed{15129}$.