How Many Clock Hand Positions Swap to a Valid Position?

At $h$ hours and $m$ minutes, the hour hand has position $(h+m/60)/12$ and the minute hand has position $m/60$ (where positions range from $0$ to $1$). If we swap positions, the new hours $h'$ and minutes $m'$ must satisfy $(h'+m'/60)/12=m/60$ and $m'/60=(h+m/60)/12$. Thus we're looking for solutions of

\begin{align} 60h'+m'&=12m\;,\\ 60h+m&=12m' \end{align}

with $0\le h,h'\lt12$, $0\le m,m'\lt60$ and $h,h'$ integer. We can solve for $m$ and $m'$ in terms of $h$ and $h'$ by elimination:

\begin{align} m&=\frac{60h+720h'}{143}\;,\\\\ m'&=\frac{60h'+720h}{143}\;. \end{align}

Then demanding $m,m'\lt60$ yields

\begin{align} h+12h'&\lt143\;,\\ h'+12h&\lt143\;. \end{align}

This is fulfilled by all combinations of $h$ and $h'$ except $h=h'=11$. Thus there are $12\cdot12-1=143$ such times, and you get them by substituting the corresonding integer values of $h,h'$ into the equations for $m,m'$ above.

A pictorial proof. Count the intersection points...