Diffusion and potential gradients for two similar atoms
This answer has been edited quite a bit. New additions are put in blockquotes.

To avoid any potential (pun not intended) confusion, let me make one thing clear before I start: Any mention of &quot;potential&quot; in this answer refers to an electric potential, not a chemical potential. The definition of electric potential can be found here on Wikipedia.


Let's start off simple and say that we have a beaker with two compartments. The two compartments are separated by a membrane that is ONLY permeable to $\ce{Na+}$ ions. In the left compartment, we put 1 M of $\ce{K+}$ ions and in the right compartment, we put 1 M of $\ce{Na+}$ ions (with appropriate counterions). If diffusion of $\ce{Na+}$ entirely ignored the presence of $\ce{K+}$ as you say, then we would expect that after a while, we would have 1 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ on the LHS, and on the RHS we would have 0.5 M $\ce{Na+}$.
This does not happen. In actual fact, there would be a concentration gradient that favours net movement of $\ce{Na+}$ from RHS-&gt;LHS, and there would be a electric potential difference that favours net movement of $\ce{Na+}$ from LHS-&gt;RHS. We could model the concentration at any point in time with:

LHS - 1 M $\ce{K+}$ and $x$ M $\ce{Na+}$
RHS - $0.5 - x$ M $\ce{Na+}$

where $x$ is between 0 and 0.5.
When $x = 0$, there is no potential difference since both sides have the same amount of charge (overall both sides are electrically neutral). However, there is a steep concentration gradient, so $\ce{Na+}$ starts diffusing from RHS-&gt; LHS, causing $x$ to increase. Now as $x$ increases, the concentration gradient becomes gentler and the potential difference starts to become larger. At one particular value of $x$, the concentration gradient will cancel out the potential difference and there will no longer be any net movement of $\ce{Na+}$ across the membrane. This is the point where we could say that diffusion has stopped.
Now, let me change the scenario and say that the membrane is permeable to BOTH $\ce{K+}$ and $\ce{Na+}$ ions. In that case, clearly the equilibrium concentrations will be 0.5 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ for both the LHS and RHS, because under these conditions there is neither a potential difference nor a concentration gradient.

It is important to note that the diffusion of $\ce{Na+}$ is NOT directly affected by the pre-existing concentration gradient of $\ce{K+}$. It is indirectly affected only because the concentration gradient of $\ce{K+}$ leads to a potential difference between the LHS and the RHS, which in turn affects the diffusion of $\ce{Na+}$. If we were to replace $\ce{K+}$ with an electrically neutral species, such as glucose, the concentration gradient of glucose would have absolutely zero effect on the diffusion of $\ce{Na+}$.



Now we can go on to explain the action potential in the neuron, which is not something I originally wanted to go into, but since the original question has been edited I feel that I have to address this now. Even though the original question now makes no reference to the action potential, it is describing exactly those events that occur during an action potential.
The inside and the outside of the neuron is separated by a cell membrane, which is selectively permeable to ions such as $\ce{Na+}$, $\ce{K+}$, $\ce{Ca^2+}$, and $\ce{Cl-}$. Of these four, sodium and potassium ions are the most important as they are directly involved in depolarisation and repolarisation when an action potential is generated.
The distribution of ions is not symmetrical across the membrane. When at rest, the extracellular concentration of $\ce{Na+}$ is much higher than the intracellular concentration of $\ce{Na+}$; the converse is true for $\ce{K+}$. Due to the membrane having different permeabilities to $\ce{Na+}$ and $\ce{K+}$ (for details of this, one can consult a neuroscience text), the resting membrane potential is -70 mV. Conventionally, the membrane potential is described as the intracellular potential with respect to the potential of the extracellular potential; this means that the inside of the cell is more negative.
Neurons react to certain stimuli by having $\ce{Na+}$ enter the cell, via various means which are not important to the current discussion. Since positively charged ions are entering the cell, this causes the membrane potential to rise. Once the membrane potential has risen to roughly -55 mV, voltage-gated $\ce{Na+}$ channels open, making the membrane permeable to $\ce{Na+}$.
At this stage, we have to consider the two factors that influence the diffusion of $\ce{Na+}$:

Since the potential difference is still negative, it favours influx of $\ce{Na+}$.
Since the extracellular concentration of $\ce{Na+}$ is larger than the intracellular concentration, the concentration gradient also favours influx of $\ce{Na+}$.

Hence more $\ce{Na+}$ enters the cell, leading to depolarisation of the membrane until the membrane potential reaches approximately +30 mV. At this stage, the voltage-gated $\ce{Na+}$ channels close and the voltage-gated $\ce{K+}$ channels open. Again, we need to consider the two factors that influence the diffusion of $\ce{K+}$:

The potential difference, which is now positive, favours efflux of $\ce{K+}$.
Since the intracellular concentration of $\ce{K+}$ is higher than the extracellular concentration, the concentration gradient also favours efflux of $\ce{K+}$.

Therefore $\ce{K+}$ leaves the cell, leading to repolarisation. There are further events such as hyperpolarisation and the refractory period, but they are not relevant to this discussion. After all this is said and done, the concentration gradients of both $\ce{Na+}$ and $\ce{K+}$ are not the same as they were at the very start of this discussion. They are restored by the $\ce{Na+}$/$\ce{K+}$ pump.
The most important thing to takeaway from this is that the diffusion of species X across a membrane is affected by two things: 1) the concentration gradient of X; 2) the potential difference across the membrane, if and only if X is a charged species. Other concentration gradients of other species, such as Y, Z etc. do not directly affect the diffusion of species X unless they contribute to the potential difference.
Prior to my edit you might have seen that I wrote that &quot;diffusion does not ignore similarly charged species&quot;. This might seem to be a contradiction with what I say now, that the concentration gradient of Y does not affect the concentration gradient of X. The answer is that the concentration gradient of Y can only affect that of X when both X and Y are charged species, and this is only an indirect effect, in that the concentration gradient of Y affects the electric potential difference which in turn affects the concentration gradient of X. If Y was uncharged, its concentration gradient would not affect the potential difference; if X was uncharged, the potential difference would not affect its concentration gradient.
I think the marble analogy is inherently flawed in that there is no real analogous concept to a potential difference. You could modify it to have some kind of analogous concept, but in my opinion it wouldn't make sense any more.