Let the smallest value of $k \in \mathbb{N}$, for which the coefficient of $x^3$ in $(1+x)^3 + (1+x)^4 + (1+x)^5 + \ldots + (1+x)^{99} + (1+kx)^{100}$, $x \neq 0$, is $\left(43n + \dfrac{101}{4}\right)\left(^{100}C_3\right)$ for some $n \in \mathbb{N}$, be $p$. Then the value of $p + n$ is:

The coefficient of $x^3$ in the given expression is the sum of the coefficients of $x^3$ in each term.

The coefficient of $x^3$ in $\sum_{r=3}^{99} (1+x)^r + (1+kx)^{100}$ is:

$\sum_{r=3}^{99} {}^{r}C_3 + k^3 {}^{100}C_3$

Using the identity $\sum_{r=k}^{n} {}^{r}C_k = {}^{n+1}C_{k+1}$, we get:

$\sum_{r=3}^{99} {}^{r}C_3 = {}^{100}C_4$

Thus, the total coefficient of $x^3$ is ${}^{100}C_4 + k^3 {}^{100}C_3$.

Given that this coefficient is equal to $\left(43n + \dfrac{101}{4}\right) {}^{100}C_3$, we have:

${}^{100}C_4 + k^3 {}^{100}C_3 = \left(43n + \dfrac{101}{4}\right) {}^{100}C_3$

Dividing both sides by ${}^{100}C_3$:

$\dfrac{^{100}C_4}{^{100}C_3} + k^3 = 43n + \dfrac{101}{4}$

Using $\dfrac{^{n}C_r}{^{n}C_{r-1}} = \dfrac{n-r+1}{r}$, we get $\dfrac{^{100}C_4}{^{100}C_3} = \dfrac{100-4+1}{4} = \dfrac{97}{4}$.

Substituting this value:

$\dfrac{97}{4} + k^3 = 43n + \dfrac{101}{4}$

$k^3 = 43n + \dfrac{101}{4} - \dfrac{97}{4}$

$k^3 = 43n + 1$

$k^3 - 1 = 43n$

Since $n \in \mathbb{N}$, we must have $n \ge 1$, which implies $k^3 - 1 \ge 43 \Rightarrow k \ge 4$.

Checking the values of $k \ge 4$ to find the smallest integer $k$ such that $k^3 - 1$ is divisible by $43$:

For $k = 4$, $k^3 - 1 = 63$ (not divisible by $43$).

For $k = 5$, $k^3 - 1 = 124$ (not divisible by $43$).

For $k = 6$, $k^3 - 1 = 215 = 43 \times 5$.

Thus, the smallest value of $k$ is $p = 6$, and the corresponding value of $n$ is $5$.

Therefore, $p + n = 6 + 5 = 11$.

Answer: $11$