Just now I think of $$$n^3$$$ idea using 2 pointers method:

1. Sort $$$a$$$
2. Choose initial pointers $$$i, j, i<j$$$, $$$a_i+a_j$$$ will be the constant sum
3. Repeat this step:
   • Increment $$$i$$$ by $$$1$$$, decrement $$$j$$$ by $$$1$$$
   • Adjust $$$i$$$ and $$$j$$$ until the constant sum is reached once again (like solution to 2SUM)

Then the answer is the max number of times step 3 is repeated across all initial $$$(i, j)$$$

Likely some optimisation is possible.

$$$1 <= N, a[i] <= 10^5$$$

$$$O(max(a)sqrt(max(a))$$$

#pragma GCC optimize("O2")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
#pragma GCC target("sse4,popcnt,abm,mmx")

#include <bits/stdc++.h>

#define out(x) cout << #x << '=' << (x) << endl
#define out2(x, y) cout << #x << '=' << (x) << ',' << #y << '=' << (y) << endl 
#define no do { cout << "No" << endl; return; } while(0)
#define yes do { cout << "Yes" << endl; return; } while (0)
#define lowbit(x) ((x) & -(x))

using namespace std;

using ll = long long;

const ll inf = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;

template<typename T> ostream & operator << (ostream &out,const set<T>&obj){out<<"set(";for(auto it=obj.begin();it!=obj.end();it++) out<<(it==obj.begin()?"":", ")<<*it;out<<")";return out;}
template<typename T1,typename T2> ostream & operator << (ostream &out,const map<T1,T2>&obj){out<<"map(";for(auto it=obj.begin();it!=obj.end();it++) out<<(it==obj.begin()?"":", ")<<it->first<<": "<<it->second;out<<")";return out;}
template<typename T1,typename T2> ostream & operator << (ostream &out,const pair<T1,T2>&obj){out<<"<"<<obj.first<<", "<<obj.second<<">";return out;}
template<typename T> ostream & operator << (ostream &out,const vector<T>&obj){out<<"vector(";for(auto it=obj.begin();it!=obj.end();it++) out<<(it==obj.begin()?"":", ")<<*it;out<<")";return out;}


const int maxn = 1e5;

const int B = sqrt(maxn) / __lg(maxn) / 2;

int cnt[maxn + 1], ans[maxn * 2 + 1];
vector<int> cntv[B + 1];

using LL = ll;

const int MAXN = 3 * 1e5 + 10, P = 998244353, G = 3, Gi = 332748118;

int limit, r[MAXN];
LL a[MAXN], b[MAXN];

inline LL fastpow(LL a, LL k) {
	LL base = 1;
	while(k) {
		if(k & 1) base = (base * a ) % P;
		a = (a * a) % P;
		k >>= 1;
	}
	return base % P;
}

inline void NTT(LL *A, int type) {
	for(int i = 0; i < limit; i++) 
		if(i < r[i]) swap(A[i], A[r[i]]);
	for(int mid = 1; mid < limit; mid <<= 1) {	
		LL Wn = fastpow( type == 1 ? G : Gi , (P - 1) / (mid << 1));
		for(int j = 0; j < limit; j += (mid << 1)) {
			LL w = 1;
			for(int k = 0; k < mid; k++, w = (w * Wn) % P) {
				 int x = A[j + k], y = w * A[j + k + mid] % P;
				 A[j + k] = (x + y) % P,
				 A[j + k + mid] = (x - y + P) % P;
			}
		}
	}
}
void ntt(int N, int M) {
    int L = 0;
    limit = 1;
	while(limit <= N + M) limit <<= 1, L++;
	for(int i = 0; i < limit; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));	
	NTT(a, 1); NTT(b, 1);
	for(int i = 0; i < limit; i++) a[i] = (a[i] * b[i]) % P;
	NTT(a, -1); NTT(b, -1);
	LL inv = fastpow(limit, P - 2);
	for(int i = 0; i <= N + M; i++) {
		a[i] = a[i] * inv % P;
		b[i] = b[i] * inv % P;
	}
}

void solve() {
    int n;
    cin >> n;
    
    for (int i = 1; i <= n; i++) {
        int a;
        cin >> a;
        cnt[a]++;
    }
    //out(B);
    //B = sqrt(maxn) / logmaxn
    //part1 maxn * sqrt(maxn)
    //part2 maxn * logmaxn ^ 2
    
    vector<int> other;
    for (int i = 1; i <= maxn; i++) {
        if (cnt[i] <= B) {
            cntv[cnt[i]].push_back(i);
        } else {
            other.push_back(i);
        }
    }
    
    //part2
    for (int x : other) {
        b[x] += 2;
        for (int y : other) {
            ans[x + y] += min(cnt[x], cnt[y]);
        }
    }
    
    //part1
    for (int i = B; i > 0; i--) {
        for (int x : cntv[i]) {
            b[x]++;
            a[x]++;
        }
        ntt(maxn, maxn);
        for (int j = 1; j <= maxn * 2; j++) {
            ans[j] += a[j] * i;
        }
        memset(a, 0, sizeof(a));
        for (int x : cntv[i]) {
            b[x]++;
        }
    }
    
    int res = 0;
    for (int i = 1; i <= maxn * 2; i++) {
        res = max(res, ans[i] / 2);
    }
    cout << res << endl;
}

int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int t = 1;
	//cin >> t;
    
    while (t--) {
    	solve(); 
	}
}