Exceptions to the Nitrogen Rule in mass spectrometry
The reason for the rule is that, outside of transition metal complexes, stable free radicals are relatively rare. If all electrons are paired and we are talking about molecules, not ions, there must also be an even number of protons. Thus an odd number of neutrons must be present to arrive at an odd molecular weight, and the only elements with the most abundant isotope having an odd number of neutrons are beryllium and nitrogen, although dysprosium is kinda close.  

In an undergraduate lab you wouldn't be given an unknown with beryllium in it because it's so dangerously toxic, so that leaves you with nitrogen unless you have a deuterated sample.  

But there are stable free radicals like nitric oxide (neurotransmitter) or chlorine dioxide (fumigant) that violate the rule. Molecular oxygen doesn't count because it's a diradical and so has an even number of electrons.

EDIT: I spoke too soon about abundances of nuclides with an odd number of neutrons. Here is a table of such nuclides with abundance $&gt; 20\%$ taken from http://atom.kaeri.re.kr/nuchart/ :
$$\begin{array}{rl} \text{Nuclide} &amp; \text{Abundance} \\ 
\hline
\sideset{^{9}}\ {Be}&amp; 100\%\\
\sideset{^{14}}\ {N}&amp; 99.636\%\\
\sideset{^{105}}\ {Pd}&amp; 22.33\%\\
\sideset{^{129}}\ {Xe}&amp; 26.4006\%\\
\sideset{^{131}}\ {Xe}&amp; 21.232\%\\
\sideset{^{163}}\ {Dy}&amp; 24.896\%\\
\sideset{^{167}}\ {Er}&amp; 22.869\%\\
\sideset{^{195}}\ {Pt}&amp; 33.78\%\\
\sideset{^{207}}\ {Pb}&amp; 22.1\%
\end{array}
$$
Of these $\sideset{^{195}}\ {Pt}$ is the most abundant isotope of $Pt$ in addition to $\sideset{^{9}}\ {Be}$ and $\sideset{^{14}}\ N$ previously mentioned.