What is the maximum number of distinct positive integer's square that sums up to $2002$? 
What is the maximum number of distinct positive integer's square that sums up to $2002$ ?

My tries:
$$\frac{n(n+1)(2n+1)}{6} = 2002$$
 $$\implies n\approx 17 $$ but am clueless as to how to proceed any further.

Your estimate gives an upper bound: Even the smallest sum obtainable from $18$ distinct positive squares is $1^2+2^2+\ldots+18^2=2109>2002$, hence we can have at most $17$ squares. 
Let's try to find a solution with $17$ squares.
If $k\le 17$ is the first number the sequence $x_1<x_2<\ldots< x_{17}$ omits, then $x_1^2+\ldots +x_{17}^2\ge 2109 -k^2$. We conclude $k\ge 11$ and want  to solve
$$ x_{11}^2+\ldots +x_{17}^2=2002-(1^2+2^2+\ldots+10^2)=1617$$
Also, $2109-k^2=2002$ has no integer solution, so that there must be at least a second omitted numbre $l$. Can we find $l>k\ge 11$ such that $1^2+2^2+\ldots+19^2-k^2-l^2=2002$? That is, $k^2+l^2=468$?
First try, $468-11^2$ is not a square. But the second try gives $468-12^2=18^2$.
So
$$2002=1^2+2^2+3^3+4^4+5^2+6^2+7^2+8^2+9^2+10^2+11^2+13^2+14^2+15^2+16^2+17^2+19^2 $$
shows that we can write $2002$ a sum of $17$ distinct positive squares.
