The answer is 13.

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Let the numbers be $x,x+1$ .

Then we have ${x}^{2}+{(x+1)}^{2}=365$

Simplifying we have :

${x}^{2}+x-182=0$

Hence we have $(x+14)(x-13)=0$

$\Rightarrow x= 13,-14$ but since negative numbers are inadmissible hence $x=13$

Finally we get the smaller number as $13$