Let $c_j$ be the number of children whose first color is $j$. Every child needs $m$ minutes of work, so the answer is at least $m$.

Take any nonempty circular interval of $k$ colors, ending at color $r$, and let $C$ children start inside it. Each of these children must use pencil $r$. After doing so, each still has at least $m-k$ lines left: reaching $r$ uses at most the first $k$ colors of their sequence.

Pencil $r$ needs $C$ different minutes for these children. Its last such use cannot finish before minute $C$, and that child needs at least $m-k$ more minutes. Thus the answer is at least $C+m-k$.

Write $x_j=c_j-1$. The bound becomes $m+\sum x_j$ over the chosen circular interval. Let $B$ be the maximum of these sums, also allowing the empty interval with sum $0$. We have proved that at least $m+B$ minutes are necessary.