Seemingly contradictory situation in electrical system loss

There is more than one relevant potential difference. You must distinguish between the potential difference, $V_L$, across the load (i.e. the 'user' that we are aiming to supply) and the potential difference, $V_W$, across just the transmitting wires (of resistance $R_W$ taken together). The load and the wires are in series across the supply so $V_text{supply}=V_L+V_W$.

The power received by the load is $IV_L$.

The power dissipated in the wires ('system loss') is $IV_W$. We can also write this as $I^2R_W$ or as $frac {V_W^2}{R_W}.$

If we increase $V_L$ by a factor of 10, we can get the same power, $IV_L$, to the load using only a tenth of the current. [The load has to be of higher resistance now, and is, in practice, the primary of a loaded step-down transformer, but that doesn't affect our argument.] So using $I^2R_W$, the power dissipated in the wires drops to $frac{1}{100}$ of its previous value. But suppose we use $frac {V_W^2}{R_W}$ ... That's fine too, because if $I$ is $frac{1}{10}$ its previous value, so is $V_W$ (since $V_W=IR_W$), so again we find that the power dissipation in the wires drops to $frac{1}{100}$ of its previous value.