The coefficient of $x^{50}$ in
\[(1 + x)^{1000} + 2x (1 + x)^{999} + 3x^2 (1 + x)^{998} + \dots + 1001x^{1000}\]can be expressed in the form $\binom{n}{k}.$  Find the smallest possible value of $n + k.$
Let
\[S = (1 + x)^{1000} + 2x (1 + x)^{999} + \dots + 1000x^{999} (1 + x) + 1001x^{1000}\]Then
\begin{align*}
xS &= x (1 + x)^{1000} + 2x^2 (1 + x)^{999} + \dots + 1000x^{1000} (1 + x) + 1001x^{1001}, \\
(1 + x)S &= (1 + x)^{1001} + 2x (1 + x)^{1000} + \dots + 1000x^{999} (1 + x)^2 + 1001x^{1000} (1 + x).
\end{align*}Subtracting these equations, we get
\[S = (1 + x)^{1001} + x(1 + x)^{1000} + \dots + x^{999} (1 + x)^2 + x^{1000} (1 + x) - 1001x^{1001}.\]Then
\begin{align*}
xS &= x(1 + x)^{1001} + x^2 (1 + x)^{1000} + \dots + x^{1000} (1 + x)^2 + x^{1001} (1 + x) - 1001x^{1002}, \\
(1 + x)S &= (1 + x)^{1002} + x (1 + x)^{1001} + \dots + x^{999} (1 + x)^3 + x^{1000} (1 + x)^2 - 1001x^{1001} (1 + x).
\end{align*}Subtracting these equations, we get
\[S = (1 + x)^{1002} - 1002x^{1001} (1 + x) + 1001x^{1002}.\]By the Binomial Theorem, the coefficient of $x^{50}$ is then $\binom{1002}{50}.$  The final answer is $1002 + 50 = \boxed{1052}.$