= Solution
Write $K_{ij}=K(\mathbf x_i,\mathbf x_j)$ and $\mathbf P=\sum_i\mathbf v_i$. The unit particle <masses> make this sum the total <momentum>. Exchange the two indices and use the <symmetry> $K_{ij}=K_{ji}$:
$$
\dot{\mathbf P}=\sum_{i,j}K_{ij}(\mathbf v_j-\mathbf v_i)
=\frac12\sum_{i,j}K_{ij}\bigl[(\mathbf v_j-\mathbf v_i)+(\mathbf v_i-\mathbf v_j)\bigr]=\boxed{\mathbf0}.
$$
Thus every interacting pair transfers equal and opposite <momentum>, even though the <velocity alignment> dissipates <kinetic energy>.
For the total <kinetic energy> $E_N=\tfrac12\sum_i|\mathbf v_i|^2$, apply the same exchange of indices:
$$
\begin{aligned}
\dot E_N
&=\sum_{i,j}K_{ij}\mathbf v_i\cdot(\mathbf v_j-\mathbf v_i)\\
&=\frac12\sum_{i,j}K_{ij}\bigl[\mathbf v_i\cdot(\mathbf v_j-\mathbf v_i)+\mathbf v_j\cdot(\mathbf v_i-\mathbf v_j)\bigr]\\
&=\boxed{-\frac12\sum_{i,j}K_{ij}|\mathbf v_j-\mathbf v_i|^2\leq0}.
\end{aligned}
$$
The dissipated <kinetic energy> is that of relative motion. Since the total <momentum> is constant, $E_N-|\mathbf P|^2/(2N)=\tfrac12\sum_i|\mathbf v_i-\mathbf P/N|^2$ also decreases. Equality holds when every pair with $K_{ij}>0$ has the same <velocity>; the hypotheses permit disconnected interacting groups, so they alone do not guarantee eventual global <velocity alignment>.
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