For how many positive integers $k$ do the lines with equations $9 x+4 y=600$ and $k x-4 y=24$ intersect at a point whose coordinates are positive integers?
Suppose that $k$ is a fixed, but unknown, positive integer.

Suppose also that the lines with equations $9 x+4 y=600$ and $k x-4 y=24$ intersect at the point with positive integer coordinates $(x, y)$.

Since $9 x+4 y=600$ and $k x-4 y=24$, adding these equations, we get $9 x+k x=624$ and so $(9+k) x=624$.

Since $x$ and $y$ are to be positive integers and $k>0$, then $9+k$ and $x$ are a positive divisor pair of 624 with $9+k>9$.

Now $624=6 \cdot 104=6 \cdot 8 \cdot 13=2^{4} 3^{1} 13^{1}$, and so the positive divisors of 624 are

$$
1,2,3,4,6,8,12,13,16,24,26,39,48,52,78,104,156,208,312,624
$$

We also want the value of $y$ to be a positive integer.

Since the point $(x, y)$ lies on the line with equation $9 x+4 y=600$, then $4 y=600-9 x$ which gives $y=150-\frac{9}{4} x$, which is an integer exactly when $x$ is a multiple of 4 .

Therefore, we want $x$ to be a positive divisor of 624 which is a multiple of 4 .

Thus, the possible values of $x$ are $4,8,12,16,24,48,52,104,156,208,312,624$.

The corresponding values of $9+k$ are $156,78,52,39,26,13,12,6,4,3,2,1$.

Since $9+k>9$, we eliminate $6,4,3,2,1$ from this list.

Thus, the possible values of $9+k$ are $156,78,52,39,26,13,12$.

The corresponding values of $k$ are 147, 69, 43, 30,17,4, 3 .

These correspond to the following values of $x: 4,8,12,16,24,48,52$.

Using $y=150-\frac{9}{4} x$, these give the following values of $y$ : 141,132,123,114,96,42,33. These are indeed all positive.

This means that there are 7 values of $k$ with the required properties.