Dan and Michael point out the relevant issues. Just for completeness, the relationship between adjusted $R^2$ and $R^2$ is given by (see, e.g., [here](https://stats.stackexchange.com/questions/342949/is-r2-adjusted-both-unbiased-and-consistent-under-the-alternative-in-simple/343113#343113))

$$
R^2_{adjusted}=1-(1-R^2)\frac{n-1}{n-2},
$$
showing that $R^2_{adjusted}=1$ if and only if $R^2=1$, which, in turn, happens when all residuals $\hat u_i=y_i-\hat y_i$ are zero, as
$$
R^2=1-\frac{\hat{u}'\hat{u}/n}{\tilde{y}'\tilde{y}/n}.
$$
Here, $\hat u$ denotes the vector of residuals and $\tilde y$ the vector of demeaned observations on the dependent variable.

Dan discusses one reason to get an $R^2$ of 1. Another is to have as many regressors as observations. So in your case, as suspected by mdewey, you maybe also only have 15 observations, in which case $R^2$ must be equal to one.

Technically, this is because the $n\times K$ (with $K$ the number of regressors) regressor matrix $X$ then is square. The OLS estimator $\hat\beta=(X'X)^{-1}X'y$ can then be written as (assuming no exact multicollinearity)
$$
\hat\beta=(X'X)^{-1}X'y=X^{-1}{X'}^{-1}X'y=X^{-1}y
$$
so that the fitted values $\hat y=X\hat\beta$ are just $\hat y=XX^{-1}y=y$, so that all residuals are zero.

Here is an illustration using artificial data (code below), in which regressors are generated totally independently of $y$, and yet we achieve an $R^2$ of 1 once we have as many of them as we have observations.

<img src="https://i.sstatic.net/6vY04.png" width="400">

Code:

    n <- 15
    regressors <- n
    y <- rnorm(n)
    X <- matrix(rnorm(regressors*n),ncol=regressors)
    
    collectionR2s <- rep(NA,regressors)
    for (i in 1:regressors){
      collectionR2s[i] <- summary(lm(y~X[,1:i]))$r.squared
    }
    plot(1:regressors,collectionR2s,col="purple",pch=19,type="b",lwd=2)
    abline(h=1, lty=2)