Report on some basic experiments. For $n \leq 30$, there is one positive eigenvalue and all the others are negative.
I checked this with the following Mathematica command:
mm[n_] := Table[Sqrt[i^2 + j^2], {i, 1, n}, {j, 1, n}]
Table[Count[Sign[Eigenvalues[SetPrecision[mm[n], 50]]], 1], {n, 1, 30}]
SetPrecision
tells Mathematica to treat the square roots as floating point numbers with $50$ decimal digits of accuracy. If you tell it to treat them as exact quantities, the computation times out; if you use the default accuracy it won't get the signs of the smallest eigenvalues right. The smallest eigenvalues here are around $10^{-30}$, so you need to be careful.
Probably the easiest way to prove this would be to exhibit an $n-1$ dimensional subspace on which this matrix is negative definite. I took my first guess, the span of the vectors $(1, -1,0,0,0,\ldots)$, $(0,1,-1,0,0,\ldots)$, $(0,0,1,-1,0,0,\ldots)$, ...
(* Change of basis matrix to the n-1 dimensional space. *)
ss[n_] := Table[If[i == j, 1, If[i == j + 1, -1, 0]], {i, 1, n}, {j, 1, n - 1}]
(* Quadratic form in the new basis. *)
qq[n_] := Transpose[ss[n]].mm[n].ss[n]
Table[PositiveDefiniteMatrixQ[SetPrecision[-qq[n], 50]], {n, 2, 30}]
For $n \leq 30$, the qudratic form is negative definite on this $n-1$ plane.