Awfully sophisticated proof for the fact :) Just to relate it with the question about Knizhnik-Zamolodchikov equation:

Find polynom p(z) with values in C[S_n] such that p'(z) = \sum_i (Id+(1i))/(z-i) p(z). [Knizhnik-Zamolodchikov equation for S_n]

Consider the following KZ ODE:

$ p'(z) = \sum_{i=2...n} \frac{ Id + \pi( (1i) )}{z-z_i} p (z) $

As it is discussed in MO-question above it is known to have polynomial solution.

The reside at infinity is equal to $Res=-\sum_{i=2...n} { Id + \pi( (1i) )}$. Which is our beloved JM-element up to sign and n*Id.

Hence its eigenvalues must be non-positive integers (this is obvious since at infinity the solution looks like $(1/z)^{Res}, so in order to be polynomial in z they must be non-positive ints). Hence we are done.

Moreover we got that eigs are greater or equal -n (as David Speyer proved directly above).

Awfully sophisticated proof for the fact :) Just to relate it with the question about Knizhnik-Zamolodchikov equation:

Find polynom p(z) with values in C[S_n] such that p'(z) = \sum_i (Id+(1i))/(z-i) p(z). [Knizhnik-Zamolodchikov equation for S_n]

Consider the following KZ ODE:

$ p'(z) = \sum_{i=2...n} \frac{ Id + \pi( (1i) )}{z-z_i} p (z) $

As it is discussed in MO-question above it is known to have polynomial solution.

The reside at infinity is equal to $Res=-\sum_{i=2...n} { Id + \pi( (1i) )}$. Which is our beloved JM-element up to sign and n*Id.

Hence its eigenvalues must be non-positive integers (this is obvious since at infinity the solution looks like $(1/z)^{Res}, so in order to be polynomial in z they must be non-positive ints). Hence we are done.

Moreover we got that eigs are greater or equal -n (as David Speyer proved directly above).