The answer is "yes" for at least one reasonable definition of "emerge", and doesn't depend on the starting weights.

For each of $k$ objects, let $X_i(t)$ be the weight of the $i$th object at time $t$. At each time step, we choose a random pair of objects $1 \leq i < j \leq k$, adding 1 to the weight of object $j$ and subtracting 1 from the weight of object $i$. Looking only at what happens to the $i$th object, we have that $$ X_i(t+1) = \begin{cases} X_i(t) & \text{if $i$ was not chosen at time $t+1$} \\ X_i(t) + 1 & \text{if $i$ and a smaller $j$ were chosen at time $t+1$} \\ X_i(t) - 1 & \text{if $i$ and a larger $j$ were chosen at time $t+1$.} \end{cases} $$

So each $X_i(t)$ is a random walk on $\mathbb Z$, with differing probabilities of moving up, down or staying still. The expectation of $X_i(t)$ is of the form $c\big(\frac{i-1}{k-1}-\frac 1 2\big)t$ for some constant $c$, so we expect the weights of the objects to be in the correct order with gaps of order $t$ between them. But the deviation from the expectation is around $\sqrt t$ with high probability by Chernoff's inequality, so typically the random fluctuations around the mean are not enough to break the ordering.

The answer is "yes" for at least one reasonable interpretation of "emerge", and doesn't depend on the starting weights.

For each of $k$ objects, let $X_i(t)$ be the weight of the $i$th object at time $t$. At each time step, we choose a random pair of objects $1 \leq i < j \leq k$, adding 1 to the weight of object $j$ and subtracting 1 from the weight of object $i$. Looking only at what happens to the $i$th object, we have that $$ X_i(t+1) = \begin{cases} X_i(t) & \text{if $i$ was not chosen at time $t+1$} \\ X_i(t) + 1 & \text{if $i$ and a smaller $j$ were chosen at time $t+1$} \\ X_i(t) - 1 & \text{if $i$ and a larger $j$ were chosen at time $t+1$.} \end{cases} $$

So each $X_i(t)$ is a random walk on $\mathbb Z$, with differing probabilities of moving up, down or staying still. The expectation of $X_i(t)$ is of the form $c\big(\frac{i-1}{k-1}-\frac 1 2\big)t$ for some constant $c$, so we expect the weights of the objects to be in the correct order with gaps of order $t$ between them. But the deviation from the expectation is around $\sqrt t$ with high probability by Chernoff's inequality, so typically the random fluctuations around the mean are not enough to break the ordering.