jmc

Let the smallest common solution be $a$. The given system of congruences is $$\begin{array}{r}a\equiv 1\equiv -1(\mathrm{mod}2),\\ a\equiv 2\equiv -1(\mathrm{mod}3),\\ a\equiv 3\equiv -1(\mathrm{mod}4),\\ a\equiv 4\equiv -1(\mathrm{mod}5).\end{array}$$ Note that if $a\equiv -1(\mathrm{mod}4)$, then $a\equiv -1(\mathrm{mod}2)$ as well, so we need only consider the final three congruences. Since $\gcd (3,4)=\gcd (4,5)=\gcd (3,5)=1$, we have $$a\equiv -1(\mathrm{mod}3\cdot 4\cdot 5),$$ that is, $a\equiv 59(\mathrm{mod}60)$.

So $a$ has a lower bound of $59$, but $59$ also happens to satisfy all of the original congruences. Thus, $a=\boxed{59}$.