How to deal with a singular Leontiev inverted matrix?
I am not sure exactly what you divided by what, but suppose your input-output table looks like this:
$$
A = 
\left[
\begin{array}{lll}
a_{11} &amp; a_{12} &amp; a_{13} \\
a_{21} &amp; a_{22} &amp; a_{23} \\
a_{31} &amp; a_{32} &amp; a_{33}
\end{array}
\right]
$$
If you proceed to divide the first column by $\sum\limits_i a_{i,1}$ the second by $\sum\limits_i a_{i,2}$ and the third by $\sum\limits_i a_{i,3}$ then your new matrix will have 1 as an eigenvalue. (You can show this by showing that the norm of any vector is unchanging upon multiplication with the matrix by straightforward calculation.) This is problematic, because if $s$ is an eigenvector then
$$
A \cdot s = s = I \cdot s 
$$
which implies
$$
(I - A) \cdot s = 0
$$
which in turn implies that $I - A$ is not invertible (singular). So my guess would be that the problem is the division.

I thought (incorrectly, see comment below) that input-output models usually deal in physical goods, not the value of physical goods, so the unit of measurement for $a_{11}$ would not be euros by say liters (of milk). In this case you would not have to divide the value of individual inputs by the sum of the value, but the number of goods used in the individual inputs by the total output of a good. This may make your problem go away.