jmc

Let the two integers be $n$ and $n+1,$ so $n(n+1)<400.$ Then the largest possible value of $n$ will be close to the square root of 400, which is $\sqrt{400}=20.$ For $n=19,$ $n(n+1)=19\cdot 20=380,$ and for $n=20,$ $n(n+1)=20\cdot 21=420,$ so the largest possible sum of two consecutive integers whose product is less than 400 is $19+20=\boxed{39}.$