Junior Balkan Mathematical Olympiad

Solution:

If we apply (i) to $l$ rows and (ii) to $k$ columns, we obtain $(m-k)l + (n-l)k$ plus signs.

a.
We have the equation $(100-k)l + (100-l)k = 2004$, or $100l + 100k - 2lk = 2004$.

$$50l + 50k - lk = 1002$$
Rewrite the last equation as
$$(50-l)(50-k) = 2500 - 1002 = 1498$$
Since $1498 = 2 \cdot 7 \cdot 107$, this equation has no solutions in natural numbers.

b.
Let $n = 101$. Then we have
$$(100-k)l + (101-l)k = 2004$$
Or
$$100l + 101k - 2lk = 2004$$
That is,
$$101k = 2004 - 100l + 2lk$$
Or
$$101k - 2lk = 2004 - 100l$$
$$k(101 - 2l) = 2004 - 100l$$
So
$$k = \frac{2004 - 100l}{101 - 2l}$$
Since $k$ must be a natural number, check possible $l$ values. For $l = 4$:
$$k = \frac{2004 - 400}{101 - 8} = \frac{1604}{93} \approx 17.25$$
Not integer. For $l = 5$:
$$k = \frac{2004 - 500}{101 - 10} = \frac{1504}{91} \approx 16.53$$
Not integer. Continue checking, but no integer solution for $n = 101$.

Let $n = 102$. Then we have
$$(100-k)l + (102-l)k = 2004$$
Or
$$100l + 102k - 2lk = 2004$$
$$50l + 51k - lk = 1002$$
Rewrite the last equation as
$$(51-l)(50-k) = 2550 - 1002 = 1548$$
Since $1548 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 43$, try $51 - l = 36$ and $50 - k = 43$. Then $l = 15$, $k = 7$.

Indeed,
$$(100 - 7) \cdot 15 + (102 - 15) \cdot 7 = 93 \cdot 15 + 87 \cdot 7 = 1395 + 609 = 2004$$
Hence, the least $n$ is $102$.