You may use $\pmod{2n+1}$.

Consider the cycle $S=1,n+1,2n+1,n,2n,n-1,\dots,n+2,1$. If $\ge n+1$ of the intervals are magical, then there are two consecutive numbers in $S$, say $a$ and $b$, whose intervals are both magical. Note that the union of $I_a$ and $I_b$ contains every number once except $b$ twice. Suppose for the sake of contradiction that there are $2m$ marked numbers, then the number of marked numbers in $I_a\cup I_b$ is $\ge 2(m+1)>2m+1$, a contradiction ($+1$ because $b$ may be marked). If the number of marked numbers is $2m+1$ yet $\le n$ intervals are magical, then there are two consecutive numbers in $S$, say $a$ and $b$, whose intervals are both not magical. Hence the number of marked numbers in $I_a\cup I_b$ is $\le 2m$, yet their union contains all $2m+1$ marked numbers, a contradiction.

We say that the interval $I_{k+n}$ goes after interval $I_k$. Overall, all intervals form a loop of length $2n+1$. Suppose $1$ holds. Then in our loop some two adjacent intervals must be magical, say $I_k$ and $I_{k+n}$. Suppose that we have $2a$ marked numbers, then each of this two intervals has at least $a+1$, the summing them up gives that in all circle plus the number $k+n$ we have $2a+2$ marked numbers, but the circle gives $2a$ and $k+n$ at most one, contradiction. Hence $1$ implies $2$. Suppose $2$ holds. Then out of any two adjacent intervals there is a magical one (because at least one has at least half of the marked elements), hence overall in the loop out of $2n+1$ intervals at least $n+1$ are magical. Hence $2$ implies $1$.