Answer: $ n = m^5 $. 
Assume that there is a partition of the set $ \{m,m+1,\dots ,m^5\} $ into two subsets $ S $ and $ T $,
neither of which contains two numbers (not necessarily distinct) along with their product. Suppose,
moreover, that $ m \in S $. The number $ m^2 = m \cdot m $ is the product of two elements of the set $ S $, so we must have
$ m^2 \in T $. From this, $ m^4= m^2\cdot m^2 \in S $. We see that $ m, m^4 \in S $, which means that the number
$ m^5= m \cdot m^4 $ belongs to the set $ T $. On the other hand, the numbers $ m, m^4 $ are elements of the set $ S $,
so the number $ m^3 $ cannot belong to this set. Therefore, $ m, m^3, m^5 \in T $.
Thus, we have $ m^2, m^3, m^5 \in T $, which is a contradiction.
On the other hand, if we divide the set $ \{m,m+1,\dots ,m^5 -1\} $ into the following two subsets: 






then the product of two elements (not necessarily distinct) of each of these subsets is not an element of that
subset. Indeed, the set $ T $ contains numbers not less than $ m^2 $, so the product of any two
of them is at least $ m^4 $, while the largest element of this set is the number $ m^4 -1 $. Meanwhile, the set $ S $ contains numbers not less than $ m $, so the product of any two is at least $ m $.
This product is less than $ m^4 $ if both numbers are less than $ m^2 $, or is at least
$ m^5 $ if one of the numbers is at least $ m^4 $, and in both cases it does not belong to the set $ S $. 
We see, therefore, that the number $ n = m^5 -1 $ does not satisfy the conditions of the problem. Even more so, numbers
$ n < m^5 - 1 $ do not satisfy these conditions, as it is sufficient to remove from each of the sets $ S, T $ formed for the value
$ m^5 - 1 $ the numbers greater than $ n $. Therefore, the sought smallest value is $ n = m^5 $.
LIX OM - I - Task 12

Given an integer $ m \geqslant 2 $. Determine the smallest such integer
$ n \geqslant m $, such that for any partition of the set $ \{m,m+1,\dots ,n\} $ into two subsets, at least one of
these subsets contains numbers $ a, b, c $ (not necessarily distinct), such that $ ab = c $.