Suppose there exists a non-trivial solution in positive integers of:
$\qquad a_1^3 + a_2^3 + a_3^3 = b_1^3 + b_2^3 + b_3^3 \qquad(1)$
with the common total < 251. Clearly we are limited to the integers $1,2,\dots,6$ since $7^3 = 343 > 251$.
We use the well-known fact that the smallest integer expressible in two ways as a sum of two cubes of positive integers is $1729$. Hence we cannot have $a_i = b_j$ for any $i,j$, otherwise the remaining two sums of cubes would be equal and the common total would be $> 1729$. (A)
The cubic residues of $1,2,\dots,6 \pmod 7$ are respectively $1,1,6,1,6,6$. The only sums that can be formed of three of these residues $\pmod 7$ are: $1 + 1 + 1 \equiv3$; $1 + 1 + 6 \equiv 1$; $1 + 6 + 6 \equiv 6$; $6 + 6 + 6 \equiv 4$. Since $3, 1, 6, 4$ are all different, any solution of $(1)$ must have the same pattern (up to order) of residues on both sides.
Case 1: Residues on both sides are ${1,1,1}$ or ${1,1,6}$
Given (A), and since the cubes of only the three integers $1,2,4$ are $\equiv 1 \pmod 7$, the cube of at least one of $1,2,4$ must be repeated on the same side of $(1)$. Also, the other side must contain two cubes of one or both of the remaining two integers in ${1,2,4}$. This yields six sub-cases to be considered:
$\qquad 1^3 + 1^3 + a^3 = 2^3 + 4^3 + b^3$ requiring $a^3 – b^3 = 70$
$\qquad 2^3 + 2^3 + a^3 = 1^3 + 4^3 + b^3$ requiring $a^3 – b^3 = 49$
$\qquad 4^3 + 4^3 + a^3 = 1^3 + 2^3 + b^3$ requiring $a^3 – b^3 = -119$
$\qquad 1^3 + 1^3 + a^3 = 2^3 + 2^3 + b^3$ requiring $a^3 – b^3 = 14$
$\qquad 2^3 + 2^3 + a^3 = 4^3 + 4^3 + b^3$ requiring $a^3 – b^3 = 112$
$\qquad 4^3 + 4^3 + a^3 = 1^3 + 1^3 + b^3$ requiring $a^3 – b^3 = -126$
The remaining two cubes $a^3, b^3$ must either be both $\equiv 1$ or both $\equiv 6 \pmod 7$. In the former sub-case (cubes of $1,2,4$) the possible values of $a^3 – b^3$ are $\pm7$, $\pm63$, $\pm56$, and in the latter sub-case (cubes of $3,5,6$) the possible values are $\pm98$, $\pm189$, $\pm91$. None of these match any of the required values, so this case is impossible.
Case 2: Residues on both sides are ${1,6,6}$ or ${6,6,6}$.
Given (A), and since the cubes of only the three integers $3,5,6$ are $\equiv 6 \pmod 7$, the cube of at least one of those integers must be repeated on the same side of $(1)$. The same method as Case 1 could be followed, but it is quicker to note that if that side has $2 \times 5^3 = 250$ or $2 \times 6^3 = 432$ the common total must be $\geq 251$. Also, if that side has $2 \times 3^3$ then the other side must contain either $2 \times 5^3$ or $2 \times 6^3$ or $5^3 + 6^3$, in each case implying a total $\geq 251$. So Case 2 cannot yield a common total $< 251$.
Hence $251$ is the lowest possible common total.
Best Answer
Slightly less technical solution. Note that the number of solutions to $x_1 + x_2 + x_3 = 100$ is just $\binom{99}{2}$ by "Stars and Bars." There are no solution with $x_1 = x_2 = x_3$. Now consider the set of solutions with exactly two of the same numbers. They are of the form $((1,1,98),\dots,(49,49,2)).$ Clearly, there are $49$ of these, each of which can be permuted in three ways. Therefore, the total number of tuples with distinct $x_1, x_2, x_3$ is just $\binom{99}{2} - 49 \cdot 3$. Each of these are counted six times by your own reasoning. Therefore, the total number of ways $100$ can be expressed as the sum of three positive integers is just $$\frac{\binom{99}{2} - 49\cdot 3}{6} = \frac{99 \cdot 49 - 3 \cdot 49}{6} = \frac{96 \cdot 49}{6} = 16 \cdot 49 = 28^2 = \fbox{784}.$$