Suppose $\mu$ is a fixed partition of $n$ of length l(\mu), and I was encountered with the following sum, namely
$\sum_{\nu} \chi_{\nu}(\mu)$.

I did some calculation using the character table that I can find (mainly Fulton & Harris's book, they have the character table up to $S_5$), and found that the sum does not vanish only if $\mu$ has an even number of even parts(someone call such $\mu$ an orthogonal partition). 

This is actually very simple to prove, only use the fact that 
$\chi_{\nu^t}(\mu)=(-1)^{n-l(\mu)} \chi_{\nu^t}(\mu)$, if $\mu$ has an odd number of even parts, then $n-l(\mu)$ is odd.

But my calculation indicates more: the sum is nonzero only if every even part of $\mu$ occurs even times.(someone also call such partition an orthogonal partition, and I donot know which is the correct definition...can anyone help?)

I checked this for $n \leq 6$ and also for $n=11$ (I found the charater table of $S_{11}$ in some paper...)

I donot know whether this is just an coincidence or this is always true. 

Since my knowledge of symmetric group is very limited, I donot hesitate to ask help on MO. Hopefully, someone will give me an answer. Thank you all!