1. Abridged Problem Statement
Given N leaflets, each requiring T_i minutes to print and then L_i minutes to deliver (deliveries start immediately after printing, with unlimited couriers). The printer handles one leaflet at a time, in any chosen order. Find the minimum possible time from start of printing until the last delivery is completed.

2. Detailed Editorial

Problem type
- Single-machine scheduling with delivery (aka “tails”).
- We have jobs i with processing time p_i=T_i and tail time q_i=L_i.
- Objective: minimize the makespan, i.e. max over i of (C_i + L_i), where C_i is its completion of printing.

Key observation (greedy order by tails)
- Consider any two jobs i and j. We ask: is it better to print i before j, or j before i?
- If we do i then j, the time to finish both deliveries is
    max( T_i+L_i,  (T_i+T_j)+L_j ).
- If we do j then i, it is
    max( T_j+L_j,  (T_j+T_i)+L_i ).
- We should pick the order that gives the smaller of these two maxima.

Proof sketch by exchange argument
- Suppose in an optimal sequence there is an adjacent pair in the wrong order by the above rule. Swapping them will not increase—and often decreases—the overall makespan. Repeating this fixes all inversions and yields the greedy schedule.

Simplified rule
- One can show this comparator is equivalent to sorting jobs in non-increasing order of L_i (i.e. largest delivery time first).

Implementation steps
1. Read N, arrays T and L.
2. Pair them up as (T_i, L_i).
3. Sort according to the two-job rule (or descending L_i).
4. Traverse the sorted list, maintain cumulative time sum of print durations.
5. After each job i, compute candidate makespan = sum + L_i; track the maximum over all i.
6. Output that maximum.

Time complexity O(N log N) for sorting, N ≤ 100 so this is very fast.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream &operator>>(istream &in, vector<T> &a) {
    for(auto &x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<pair<int, int>> a;

void read() {
    cin >> n;
    a.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i].first;
    }
    for(int i = 0; i < n; i++) {
        cin >> a[i].second;
    }
}

void solve() {
    // Current time: max(a1 + b1, a1 + a2 + b2)
    // Alternative time: max(a2 + b2, a2 + a1 + b1)
    //      -= a1 + a2
    // Current time: max(b1 - a2, b2)
    // Alternative time: max(b2 - a1, b1)
    //       If b1 - a2 > b2
    //             Current time: b1 - a2
    //             Alternative time: b1         -> Alternative time is longer
    //       If b1 - a2 < b2
    //             Current time: b2
    //             Alternative time: max(b2 - a1, b1)
    //                                          -> Alternative time is better
    //                                          only
    //                                             if b1 < b2

    // Current time: max(a1 + b1, a1 + a2 + b2)
    // Alternative time: max(a2 + b2, a2 + a1 + b1)
    // -> b1 < b2
    //    Current time: a1 + b1 + b2
    //    Alternative time: max(a1 + a2 + b1, a2 + b2) < a1 + b1 + b2
    // -> b1 > b2
    //    Current time: max(a1 + b1, a1 + a2 + b2)
    //    Alternative time: a1 + a2 + b1 > Current time

    stable_sort(
        a.begin(), a.end(),
        [](const pair<int, int> &x, const pair<int, int> &y) {
            return max(x.first + x.second, x.first + y.first + y.second) <
                   max(y.first + y.second, y.first + x.first + x.second);
        }
    );

    int ans = 0, sum = 0;
    for(int i = 0; i < n; i++) {
        sum += a[i].first;
        ans = max(ans, sum + a[i].second);
    }

    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution
```python
import sys

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    n = int(next(it))
    # Read T_i and L_i
    T = [int(next(it)) for _ in range(n)]
    L = [int(next(it)) for _ in range(n)]
    # Pair them as (T_i, L_i)
    jobs = list(zip(T, L))
    # Sort by the two-job comparator:
    # For speed, we can just sort descending by L_i,
    # which is equivalent to the pairwise rule.
    jobs.sort(key=lambda x: -x[1])

    current_time = 0  # cumulative printing time
    answer = 0        # maximum C_i + L_i

    for t, l in jobs:
        current_time += t       # finish printing this job
        answer = max(answer, current_time + l)
    print(answer)

if __name__ == "__main__":
    main()
```

Comments on the sorting key
- Sorting by `-L_i` (i.e. largest delivery time first) implements the same optimal order as the pairwise max-makespan comparator.

5. Compressed Editorial
- Model: single-machine scheduling with tails (print times p_i, delivery q_i).
- Goal: minimize max_i (C_i + q_i).
- Greedy rule: order jobs so that for any pair (i,j) placing i before j yields smaller max than j before i.
- This reduces to sorting by descending q_i.
- Simulate cumulative print time, track the maximum of (cumulative + q_i).
