jmc

Let's find the sum of the first ten positive integers: $1+2+\dots +10=\frac{10(1+10)}{2}=55$. Now we need to find which of the first ten positive integers, when subtracted from 55, gives a perfect square, namely 49 (36 is 19 away from 55, so it cannot be obtained). The desired integer is $\boxed{6}$ since $55-6=49=7^2$.