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 sumT of print durations.  
5. After each job i, compute candidate makespan = sumT + 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. Provided C++ Solution with Detailed Comments  

#include <bits/stdc++.h>  
using namespace std;

// Overload << to print a pair as "first second"
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {
    return out << x.first << ' ' << x.second;
}

// Overload >> to read into a pair from two space-separated values
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &x) {
    return in >> x.first >> x.second;
}

// Overload >> to read all elements of a vector from input
template<typename T>
istream &operator>>(istream &in, vector<T> &a) {
    for (auto &x: a) {
        in >> x;
    }
    return in;
}

// Overload << to print all elements of a vector, separated by spaces
template<typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
    for (auto x: a) {
        out << x << ' ';
    }
    return out;
}

int n;  // Number of leaflets
vector<pair<int,int>> a;  // For each leaflet: (print_time T, delivery_time L)

// Read input: n, then all T_i, then all L_i
void read() {
    cin >> n;
    a.resize(n);
    // read T_i into a[i].first
    for(int i = 0; i < n; i++) {
        cin >> a[i].first;
    }
    // read L_i into a[i].second
    for(int i = 0; i < n; i++) {
        cin >> a[i].second;
    }
}

// Solve one test case
void solve() {
    // We sort leaflets by the rule that minimizes the maximum completion+delivery when comparing any two.
    stable_sort(
        a.begin(), a.end(),
        [](const pair<int, int> &x, const pair<int, int> &y) {
            // Compute makespan if x goes before y
            int xy = max(x.first + x.second, x.first + y.first + y.second);
            // Compute makespan if y goes before x
            int yx = max(y.first + y.second, y.first + x.first + x.second);
            // Pick the ordering that yields the smaller makespan
            return xy < yx;
        }
    );

    int ans = 0;    // Will hold the minimal possible overall time
    int sum = 0;    // Cumulative printing time so far

    // Simulate printing in the chosen order
    for (int i = 0; i < n; i++) {
        sum += a[i].first;               // finish printing leaflet i at time 'sum'
        ans = max(ans, sum + a[i].second); // delivery of i finishes at sum + L_i
    }

    cout << ans << '\n';  // Output the result
}

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

    int T = 1;  // single test
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}

4. Python Solution with Detailed Comments  

```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).