We show bijectively that
$$2^2+4^2+6^2+\cdots +(2n)^2=\binom{2n+2}{3}.$$
That does not quite give a purely combinatorial expression for $1^2+2^2+ \cdots +n^2$, since we still need to divide by $4$, which is "algebra." But the sin of multiplying or dividing by a constant seems to be a small one in this game. And the result is interesting.
Imagine choosing $3$ numbers from the numbers $1, 2, \dots, 2n+2$. This can be done in $\binom{2n+2}{3}$ ways.
We will show that the number of choices is also
$$2^2+4^2+6^2+\cdots +(2n)^2.$$
Maybe the largest number chosen is $2k+1$ or $2k+2$. If it is $2k+1$, the other two can be chosen in $\binom{2k}{2}$ ways, and if it is $2k+2$, then the other two can be chosen in $\binom{2k+1}{2}$ ways. Using "algebra", we find that
$$\binom{2k}{2}+\binom{2k+1}{2}= (2k)^2.$$
Take $k=1, 2, 3, \dots, n$. We conclude that the number of ways of choosing $3$ numbers from $1, 2, \dots, 2n+2$ is
$$2^2+4^2+6^2+\cdots +(2n)^2.$$
Since the number of choices is clearly $\binom{2n+1}{3}$, this almost completes the argument. (In a note at the end of this post, we deal with the minor task of eliminating the use of algebra.)
Comment: If we allow division by $4$, we can conclude that
$$1^2+2^2+\cdots +n^2=\frac{1}{4}\binom{2n+2}{3}.$$
I used to think that in the usual expression $\frac{(n)(n+1)(2n+1)}{6}$ for the sum of the first $n$ squares, the $2n+1$ somehow is the weird term, it does not fit in. But it does! Actually, it is $n$ and $n+1$ that are strange. We can think of $(n)(n+1)(2n+1)$ as ``really'' being $(1/4)[(2n)(2n+1)(2n+2)]$.
Cleaning up: There is a standard way to avoid the ``calculation''
$$\binom{2k}{2}+\binom{2k+1}{2}= (2k)^2.$$
For completeness we show bijectively that for any $m$, in particular $m=2k$, we have
$$\binom{m}{2} +\binom{m+1}{2}=m^2.$$
For let $M$ be the collection $\left\{1,2,3,\dots,m\right\}$. Then the number of ordered pairs $(a,b)$ with $a$ and $b$ in $M$ is $m^2$. These ordered pairs are of two types: (i) the ones with $a<b$ and (ii) the ones with $a \ge b$. The number of ordered pairs $(a,b)$ with $a<b$ is just $\binom{m}{2}$. The ordered pairs $(a,b)$ with $a \ge b$ are in a natural bijection with the ordered pairs $(b,a+1)$ with $b<a+1$, and there are $\binom{m+1}{2}$ of these.