Firstly, there are $12$ unit squares which contain an even number. Every $2\times2$ square in the diagram has entries which consist of two odd numbers and two even numbers and hence have an even total. There are $16$ of these. Each $3\times3$ square in the diagram, however, has entries which consist of five odd numbers and four even numbers (giving an odd total), or four odd numbers and five even numbers (giving an even total). There are $4$ of the latter: those with $8$, $12$, $14$ or $18$ in the centre. Every $4\times4$ square in the diagram has entries which consist of eight odd numbers and eight even numbers and hence have an even total. There are $4$ of these. Finally, the full $5\times5$ square contains $13$ odd numbers and $12$ even numbers, giving an odd total. So the required number is $12+6+4+4$, that is $36$.

*This problem is taken from the UKMT Mathematical Challenges.*

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