Estonian Mathematical Olympiad

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.