#include <bits/stdc++.h>
using namespace std; 
 
#define fst first
#define snd second 
 
typedef long long ll; 
typedef pair<int, int> ii; 
 
const ll LINF = (ll)1e18; 
const int INF = (int)1e9;
 
const int N = (int)2e5 + 5; 
 
// Đếm số cặp (i, j) (i < j) sao cho a[i] + a[j] > b[i] + b[j] 
//                               <=> a[i] - b[i] > b[j] - a[j]
// Nên ở đây các em có thể tạo 1 mảng c chứa các giá trị b[j] - a[j], sort c tăng dần
// Sau đó với mỗi i ở mảng a thì ta đếm có bao nhiêu j ở c thoả mãn c[j] < a[i] - b[i]
 
int n, a[N], b[N];
 
int main() {
	ios::sync_with_stdio(0); 
	cin.tie(0); 
	cin >> n; 
	for (int i = 1; i <= n; i++) cin >> a[i]; 
 
	vector<int> c; 
	for (int i = 1; i <= n; i++) {
		cin >> b[i];
		c.push_back(b[i] - a[i]);  
	}
 
	sort(c.begin(), c.end()); 
 
	ll cnt = 0; 
	for (int i = 1; i <= n; i++) {
		//    vị trí j lớn nhất sao cho c[j] < x
		// = (vị trí j' đầu tiên sao cho c[j] >= x) - 1
		cnt += (lower_bound(c.begin(), c.end(), a[i] - b[i]) - c.begin() - 1) + 1; // vì mảng c đánh số từ 0  
		cnt -= (b[i] - a[i] < a[i] - b[i]); // nhớ trừ trường hợp i = j 
	}
 
	cout << cnt / 2 << '\n'; // Mỗi cặp (i, j) bị đếm lặp 2 lần
}

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5
4 8 2 6 2
4 5 4 1 3