Answer: No.

The numbers $1950$, $1934$, and $2024$ give remainders $4$, $2$, and $1$, respectively, when divided by $7$. Raising $4$ to powers $n = 1, 2, 3, 4, 5, 6, \dots$ results in remainders $4, 2, 1, 4, 2, 1, \dots$, and raising $2$ to the same powers results in remainders $2, 4, 1, 2, 4, 1, \dots$. Thus, the remainders of the left side of the given equation for powers $n = 1, 2, 3, 4, 5, 6, \dots$ are $6, 6, 2, 6, 6, 2, \dots$. The remainder of the right side is $1$ for every $n$. The contradiction shows that there is no positive integer $n$ that satisfies the given equation.

Clearly $n = 1$ is not suitable, because $1950 + 1934 > 2024$. It is also easy to verify that

$$1950^2 + 1934^2 = 3802500 + 3740356 = 7542856 > 4096576 = 2024^2.$$

For $n \ge 3$, there are no integer solutions to the equation $a^n + b^n = c^n$ by Fermat's Last Theorem.