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The correct answer is $289$. A quadratic equation of the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has no real solutions when the value of the discriminant, $b^2-4ac$, is less than $0$. In the given equation, $x^2-34x+c=0$, $a=1$ and $b=-34$. Therefore, the discriminant of the given equation can be expressed as ${\left(-34\right)}^2-4\left(1\right)\left(c\right)$, or $\mathrm{1,156}-4c$. It follows that the given equation has no real solutions when $\mathrm{1,156}-4c<0$. Adding $4c$ to both sides of this inequality yields $\mathrm{1,156}<4c$. Dividing both sides of this inequality by $4$ yields $289<c$, or $c>289$. It's given that the equation $x^2-34x+c=0$ has no real solutions when $c>n$. Therefore, the least possible value of $n$ is $289$.