Minimum of the smallest

Letting $x = 2m$ for some positive integer $m,$ we require that

$2m + 2(m + 1) + 2(m + 2) + ..... + 2(m + (n - 1)) = 154$

$\Longrightarrow m + (m + 1) + (m + 2) + ... + (m + (n - 1)) = 77$

$\Longrightarrow \dfrac{(m + n - 1)(m + n)}{2} - \dfrac{(m - 1)m}{2} = 77$

$\Longrightarrow m^{2} + 2mn + n^{2} - m - n - m^{2} + m = 154$

$\Longrightarrow n(n + 2m - 1) = 154.$

So in order to minimize $m$ we will want $n \lt (n + 2m - 1)$ as close together as possible such that their product is $154.$

Now since $154 = 2*7*11,$ we can factor it as a product of two ascending terms as $1*154 = 2*77 = 7*22 = 11*14,$ so $m$ will be minimized when $n = 11$ and $n + 2m - 1 = 14 \Longrightarrow m = 2.$ This results in a smallest possible value for $x$ of $2m = \boxed{4}.$

Comments: Note that the fact that $\sum_{i=1}^{k} i = \frac{k(k + 1)}{2}$ was used in this proof. Note also that the other possible solutions $(x,n)$ are $(16,7), (76,2)$ and $(154,1).$

The problem is equivalent to writing 77 as a sum of consecutive positive integers.

Now, for every odd factor $m$ of 77, we can write 77 as the sum of $m$ consecutive integers (not necessarily positive) with middle value $77/m$ , and all representations with an odd number of summands are of this form.

$77=77=8+9+10+11+12+13+14=2+3+...+7+..+11+12$ $=-37-36-...+1+...+36+37+38+39$

The last sum simplifies to $77=38+39$ .

By doubling everything, we get the representations of 154 we seek, and the smallest initial term is $\boxed{4}$ .