, 1998

Solution:

Let this smallest positive integer be represented as $a^{3}+b^{3}+c^{3}=d^{3}+e^{3}+f^{3}$. By inspection, a solution is not possible with the first 4 cubes. We prove that it is impossible to write the same number as two different sums of the first 5 cubes. Because we necessarily need to use the 5th cube (otherwise, this proof would be for the first 4 cubes), we have $5^{3}+b^{3}+c^{3}=d^{3}+e^{3}+f^{3}$. Without loss of generality, suppose $d=5$. By inspection, there is no solution to $b^{3}+c^{3}=e^{3}+f^{3}$, such that $b, c, e, f \leq 5$ and $b, c$ and $e, f$ are unique.
Then none of $d, e, f$ are $5$. Then at least two must be $4$, otherwise the RHS would be too small. Without loss of generality, suppose $d=e=4$. Then $b^{3}+c^{3}=3+f^{3}$. By inspection, there are no possible solutions if $b, c, f \leq 4$.
Thus if $a=5$, there are no solutions.
Suppose that there is a solution within the first 6 cubes. Then $a=6$. By the same analysis as above, $d=e=5$, otherwise the RHS would be too small. Then $b^{3}+c^{3}=34+f^{3}$. By inspection, we see that a possible solution is $b=3, c=2, f=1$. Then the desired integer is $6^{3}+3^{3}+2^{3}=251$.