Hint: Test a few values of $n$ to get the idea. (Don't actually compute the sum, check what numbers you are summing). Try $2, 99,3,98$ and perhaps $50$ if you still don't get what's going on after trying those.

$\frac{101}{2}$ is not the only solution. All real numbers $x$ such that $50\leq x \leq 51$ work.

No, $\frac{101}{2}$ isn't even a solution, and just assuming stuff like that really is like what middle school math does. For rigor, consider the graph from the intervals $[1,2], [2,3]\cdots, [99, 100]$, and then it is quite evident that the graph is somewhat like a upwards opening parabola symmetric about the line $x=\frac{101}{2}$, so we minimum is obviously at $x=50, 51$.

math42 wrote: No, $\frac{101}{2}$ isn't even a solution, and just assuming stuff like that really is like what middle school math does. For rigor, consider the graph from the intervals $[1,2], [2,3]\cdots, [99, 100]$, and then it is quite evident that the graph is somewhat like a upwards opening parabola symmetric about the line $x=\frac{101}{2}$, so we minimum is obviously at $x=50, 51$. Wolfram Alpha says otherwise. (Unless I'm very very very bad)

@Darn, math42, worstindium You all have the right idea, but notice that it's asking for which integers this is minimized at. Of course, it is minimized at the real number $\frac{101}{2}$, but then we want the minimum integers!

math42 wrote: No, $\frac{101}{2}$ isn't even a solution, and just assuming stuff like that really is like what middle school math does. Dang, thats harsh

Darn I should actually read the problem before posting ._.