Special subdivision of numbers from 1 to 99 I've been lately working on a problem I still can't solve. The problem is:
Can we divide numbers from 1 to 99 into 33 groups of three numbers, such that in every group one number is the sum of the two remaining elements?
Thank you in advance for any help or indication :)
 A: Not an answer, but some numerical data, generated using the following quick code:
#include <vector>
#include <iostream>
#include <cstdlib>

using namespace std;

void getCandidates(int sum, const vector<bool> &nums, vector<int> &choices)
{
    choices.clear();
    for(int i=0; i<(sum-1)/2; i++)
    {
        if(nums[i] && nums[sum-i-2] && i != sum-i-2)
        {
            choices.push_back(i);
        }
    }
}

bool recurse(const vector<bool> &nums)
{
    int max=-1;
    for(int i=0; i<nums.size(); i++)
    {
        if(nums[i])
            max = i;
    }
    if(max == -1)
    {
        return true;
    }
    vector<int> cands;
    getCandidates(max+1, nums, cands);
    for(int i=0; i<cands.size(); i++)
    {
        vector<bool> copy = nums;
        copy[max] = false;
        copy[cands[i]] = false;
        copy[max-cands[i]-1] = false;
        if(recurse(copy))
            return true;
    }
    return false;
}

void testSize(int size)
{
    vector<bool> nums;
    for(int i=0; i<size; i++)
        nums.push_back(true);
    cout << size << ": ";
    if(recurse(nums))
        cout << "yes";
    else
        cout << "no";
    cout << endl;
}

int main()
{
    for(int i=0; i<33; i++)
        testSize(3*i);
    return 0;
}

Output so far is pasted below. Worst-case runtime is exponential, so it may never get to 99, but I'll keep this updated as the program keeps running.
0: yes
3: yes
6: no
9: no
12: yes
15: yes
18: no
21: no
24: yes
27: yes
30: no
33: no
36: yes
39: yes
42: no
45: no
48: yes
51: yes
54: no
57: no
60: yes
63: yes

So far the answer seems to be "yes" whenever $n$ satisfies the necessary condition that $\frac{n(n+1)}{2}$ is even.
EDIT: Here is the solution for 24 as requested by Leen above.
(14, 4, 10) (15, 3, 12) (17, 8, 9) (18, 7, 11) (19, 6, 13) (21, 5, 16) (22, 2, 20) (24, 1, 23)
A: If it works for $n$ then it works for $4n$ and $4n+3$.
This can be seen as follows:
assume it works for $n$, then it works for the even numbers until $2n$ (just multiply everything by 2).
Now we take following combinations:
$(2n+1)+(2n+2)=4n+3$, $(2n-1)+(2n+3)=4n+2$, $(2n-3)+(2n+4)=4n+1$, $\ldots$, $1+(3n+2)=3n+3$.
The only numbers we did not use are the even numbers until $2n$.
An analogous argument shows that it works for $4n$ if it works for $n$.
user7530 has shown that it works for $n=24$ in another answer, so we have an affirmative answer, since $99=4*24+3$.
(Thanks for showing the error in my previous answer, Dennis. I hope this one makes more sense.)
A: Generalise $n$ and let $n=3m$. This problem is a laxer version of the following more restrictive one:
Partition $\{1, \dots, 3m\}$ into $m$ 3-sets $\{x_i, y_i, z_i\}$, $i=1, \dots, m$, where $x_i+y_i=z_i$ and $y_i\leqslant m$.
This more restrictive problem is equivalent to one studied by Skolem and another studied by Nickerson. Nickerson's problem is in turn a variant of a problem studied by Langford. Nowakowski, ch.1, states these problems, shows equivalences among them, and also gives solutions, for which he cites Davies and Skolem (the latter two citations are taken from 
 the bibliography in Nowakowski). 
R. J. Nowakowski. Generalizations of the Langford-Skolem problem. MS Thesis, Dept. Math., Univ. Calgary, May 1975.
R.O. Davies. On Langford's Problem. II, Math. Gaz. 43 (1959), pp.253-255.
T. Skolem. On certain distributions of integers in pairs with given differences. Math. Scand. 5 (1957), pp.57-58; M.R. 19, 1159.
