How many ordered pairs $(s, d)$ of positive integers with $4 \leq s \leq d \leq 2019$ are there such that when $s$ silver balls and $d$ diamond balls are randomly arranged in a row, the probability that the balls on each end have the same color is $\frac{1}{2}$?

$ \mathrm{(A) \ } 58 \qquad \mathrm{(B) \ } 59 \qquad \mathrm {(C) \ } 60 \qquad \mathrm{(D) \ } 61 \qquad \mathrm{(E) \ } 62$
1. We start by analyzing the given condition: the probability that the balls on each end have the same color is $\frac{1}{2}$. This implies that the number of ways to arrange the balls such that the ends are the same color is equal to the number of ways to arrange the balls such that the ends are different colors.

2. Let $s$ be the number of silver balls and $d$ be the number of diamond balls. The total number of balls is $s + d$. The total number of ways to arrange these balls is $\binom{s+d}{s}$.

3. The number of ways to arrange the balls such that the ends are the same color can be split into two cases:
   - Both ends are silver: $\binom{s+d-2}{s-2}$
   - Both ends are diamond: $\binom{s+d-2}{d-2}$

4. The number of ways to arrange the balls such that the ends are different colors is the total number of arrangements minus the number of ways the ends are the same color:
   \[
   \binom{s+d}{s} - \left( \binom{s+d-2}{s-2} + \binom{s+d-2}{d-2} \right)
   \]

5. Given that the probability of the ends being the same color is $\frac{1}{2}$, we have:
   \[
   \frac{\binom{s+d-2}{s-2} + \binom{s+d-2}{d-2}}{\binom{s+d}{s}} = \frac{1}{2}
   \]

6. Simplifying the equation, we get:
   \[
   \binom{s+d}{s} = 2 \left( \binom{s+d-2}{s-2} + \binom{s+d-2}{d-2} \right)
   \]

7. Using the definition of binomial coefficients, we can rewrite the equation as:
   \[
   \frac{(s+d)!}{s!d!} = 2 \left( \frac{(s+d-2)!}{(s-2)!d!} + \frac{(s+d-2)!}{s!(d-2)!} \right)
   \]

8. Simplifying further, we get:
   \[
   (s+d)! = 2 \left( (s+d-2)! \left( \frac{s(s-1)}{d!} + \frac{d(d-1)}{s!} \right) \right)
   \]

9. Canceling out the common factorial terms, we obtain:
   \[
   (s+d)(s+d-1) = 2(s^2 - s + d^2 - d)
   \]

10. Rearranging terms, we get:
    \[
    2sd = s^2 - s + d^2 - d
    \]

11. This can be rewritten as:
    \[
    s + d = (s - d)^2
    \]

12. We now need to find pairs $(s, d)$ such that $4 \leq s \leq d \leq 2019$ and $s + d = (s - d)^2$. We realize that $s$ and $d$ must be consecutive triangular numbers.

13. The smallest values are $\frac{3 \cdot 4}{2} = 6$ and $\frac{4 \cdot 5}{2} = 10$, while the largest values are $\frac{62 \cdot 63}{2} = 1953$ and $\frac{63 \cdot 64}{2} = 2016$.

14. Counting the number of such pairs, we find there are 60 pairs.

The final answer is $\boxed{60}$