It's a theoretical demand :
$$
 \begin{pmatrix}
  \nu_{e}\\
  \nu_{\mu}\\
  \nu_{\tau}
 \end{pmatrix}
 =
 \begin{pmatrix}
  U_{e1} & U_{e2} & U_{e3} \\
  U_{\mu1} & U_{\mu2} & U_{\mu3}  \\
  U_{\tau1} & U_{\tau2} & U_{\tau3}
 \end{pmatrix}
 \begin{pmatrix}
  \nu_{1}\\
  \nu_{2}\\
  \nu_{3}
 \end{pmatrix}
$$

You know that all states are normalized, for example  :
$⟨\nu_{e}|\nu_{e}⟩=1=(U_{e1}^{*}⟨\nu_{1}|+U^{*}_{e2}⟨\nu_{2}|+U^{*}_{e3}⟨\nu_{3}|)( U_{e1}|\nu_{1}⟩+U_{e2}|\nu_{2}⟩+U_{e3}|\nu_{3}⟩ )$

so 

$U_{e1}^{*}U_{e1}+U_{e2}^{*}U_{e2}+U_{e3}^{*}U_{e3}=1$

You can do the same for the whole matrix and find $U^{+}U=I$


EDIT : as dmckee pointed out it's a general feature in quantum mechanics, the matrix you use to change the basis (here from mass eigenstate to flavour eigenstate) must be unitary.