As others suggested, I am posting my earlier comment as an answer:

Sure. Let $n$ be a positive integer. Let $N$ be the product of $2n$ and Euler $\phi(2n)$. Then for each prime $p$ that divides $n$, the number $p-1$ divides $N$. So for each prime divisor $p$ of $n$, the denominator of the Bernoulli number $B_N$ is div'l by $p$. So none of the prime factors of $n$ cancel with factors in the numerator of $B_N/N$. So $denom(B_N/N)$ is divisible by $n$. Now $denom(B_N/N)$ or $2*denom(B_N/N)$ is the order of a cyclic summand in the image of the J-homomorphism in the $(2N-1)$st stable homotopy group of $S^0$. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order $n$.

As others suggested, I am posting my earlier comment as an answer:

$\DeclareMathOperator\denom{denom}$Sure. Let $n$ be a positive integer. Let $N$ be the product of $2n$ and Euler $\phi(2n)$. Then for each prime $p$ that divides $n$, the number $p-1$ divides $N$. So for each prime divisor $p$ of $n$, the denominator of the Bernoulli number $B_N$ is divisible by $p$. So none of the prime factors of $n$ cancel with factors in the numerator of $B_N/N$. So $\denom(B_N/N)$ is divisible by $n$. Now $\denom(B_N/N)$ or $2\denom(B_N/N)$ is the order of a cyclic summand in the image of the J-homomorphism in the $(2N-1)$st stable homotopy group of $S^0$. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order $n$.